Students can find the solutions of the Problems In General Physics I.E. Irodov - Kinematics on this page. The branch of physics that deals with motion with respect to space and time without considering the cause of that motion, is known as kinematics. This is an important topic for the JEE Main exam. The important topics in kinematics include particle, motion, displacement and distance, average speed and average velocity, etc. It also includes velocity and acceleration of a point, average vectors of velocity, and acceleration of a point, distance covered by a point, angular velocity and angular acceleration of a solid body, acceleration of a point expressed in projections on the tangent, and the normal to a trajectory, etc.
By referring to these solutions, students can achieve higher marks in the JEE exam. Students are recommended to go through these solutions thoroughly before the exams to score well and improve their problem-solving abilities.
1. A motorboat going downstream overcame a raft at a point A; T = 60 min later it turned back and after some time passed the raft at a distance l = 6.0 km from the point A. Find the flow velocity assuming the duty of the engine to be constant.
Solution:
Let v_{0} be the stream velocity and v’ the velocity of motorboat w.r.t. water. The motorboat reached point B while going downstream with velocity (v_{0} + v’) and then returned with velocity (v’ - v_{0}) and passed the raft at point C. Let t be the time for the raft (which flows with stream velocity v_{0}) to move from point A to C, during which the motorboat moves from A to B and then from B to C.
Therefore l/v_{0} = T + ((v_{0} + v’)T - l)/(v’ - v_{0})
On solving we get v_{0} = l/2T
2. A point traversed half the distance with a velocity v_{0}. The remaining part of the distance was covered with velocity v_{1} for half the time, and with velocity v_{2} for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
Solution:
Let s be the total distance traversed by the point and t_{1} the time taken to cover half the distance. Further, let 2t be the time to cover the rest half of the distance.
Therefore s/2 = v_{0}t_{1}
or t_{1} = s/2v_{0 }…(1)
And s/2 = (v_{1} + v_{2})t
Or 2t = s/(v_{1} + v_{2}) …(2)
Hence the sought average velocity
<v> = s/t_{1} + 2t
= s/([s/2v_{0}] + [s/(v_{1} + v_{2})])
= 2v_{0}(v_{1} + v_{2})/(v_{1} + v_{2} + 2v_{0})
3. A car starts moving rectilinearly, first with acceleration w = 5.0 m/s^{2} (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate w, comes to a stop. The total time of motion equals t = 25 s. The average velocity during that time is equal to (v) = 72 km per hour. How long does the car move uniformly?
Solution:
As the car starts from rest and finally comes to a stop, and the rate of acceleration and deceleration are equal, the distances, as well as the times taken, are the same in these phases of motion. Let ∆t be the time for which the car moves uniformly. Then the acceleration/ deceleration time is (t - ∆t)/2 each. So,
v t = 2{½ w (t - ∆t)^{2}/4} + w(t - ∆t) ∆t/2
∆t^{2} = t^{2} - (4vt/w)
Hence ∆t = t√(1 - 4v/wt) = 15 s
4. A point moves rectilinearly in one direction. Fig. 1.1 shows the distance s traversed by the point as a function of the time t.
Using the plot find:
(a) the average velocity of the point during the time of motion;
(b) the maximum velocity;
(c) the time moment to at which the instantaneous velocity is equal to the mean velocity averaged over the first to seconds.
Solution:
(a) Sought average velocity
v = s/t = 200cm/20s
= 10cm/s
(b) For the maximum velocity, ds/dt should be maximum. From the figure ds/dt is maximum for all points on the line ac, thus the sought maximum velocity becomes average velocity for the line ac and is equal to :
bc/ab = 100cm/4s = 25 cm/s
(c) Time t_{0} should be such that corresponding to it the slope ds/dt should pass through the point O (origin), to satisfy the relationship ds/dt = s/t_{0}. From figure the tangent at point d passes through the origin and thus corresponding time t = t_{0} = 16 s
5. Two particles, 1 and 2, move with constant velocities v_{1} and v_{2}. At the initial moment, their radius vectors are equal to r_{1} and r_{2}. How must these four vectors be interrelated for the particles to collide?
Solution:
Let the particles collide at the point A (Fig.), whose position vector is
(say).If t be the time taken by each particle to reach at point A, from triangle law of vector addition :
So
Therefore
From equations (1) and (2)
Or
which is the sought relationship.
6. A ship moves along the equator to the east with velocity v_{o} = 30 km/hour. The southeastern wind blows at an angle φ = 60^{0 }to the equator with velocity v = 15 km/hour. Find the wind velocity v' relative to the ship and the angle φ' between the equator and the wind direction in the reference frame fixed to the ship.
Solution:
We have
From the vector diagram of eq (1) and using the properties of triangle
v’ = √(v_{0}^{2} + v^{2} + 2v_{0}v cos φ) = 39.7 km/hr (2)
And v’/sin (π - φ) = v/sin θ
Or sin θ = v sin φ/v’
θ = sin^{-1}(v sin φ/v’)
Using (2) and putting the values of v and d
θ = 19.1^{0}
7. Two swimmers leave point A on one bank of the river to reach point B lying right across on the other bank. One of them crosses the river along the straight line AB while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get to point B. What was the velocity u of his walking if both swimmers reached the destination simultaneously? The stream velocity v_{0} = 2.0 km/hour and the velocity v’ of each swimmer with respect to water equals 2.5 km per hour.
Solution:
Let one of the swimmer (say 1) cross the river along AB, which is obviously the shortest path. Time taken to cross the river by the swimmer 1.
t_{1} = d/√(v’^{2}- v_{0}^{2}), (where AB = d is the width of the river.) ..(1)
For the other swimmer (say 2), which follows the quickest path, the time taken to cross the river.
t_{2} = d/v’
In the time t_{2} drifting of the swimmer 2 becomes
x = v_{0}t_{2} = v_{0}d/v’ (using eq. 2) ..(3)
If t_{3} be the time for swimmer 2 to walk the distance x to come from C to B, then
t_{3} = x/u = v_{0}d/v’u (using eq. 3) ..(4)
according to the problem t_{1} = t_{2} + t_{3}
Or d/√(v’^{2} - v_{0}^{2}) = d/v’ + v_{0}d/v’u
On solving we get u = v_{0}/((1 - v_{0}^{2})/v’^{2})^{-1/2} - 1
= 3 km/hr
8. Two boats, A and B, move away from a buoy anchored at the middle of a river along the mutually perpendicular straight lines: the boat A along the river, and the boat B across the river. Having moved off an equal distance from the buoy the boats returned. Find the ratio of times of motion of boats T_{A}/T_{B} if the velocity of each boat with respect to water is η = 1.2 times greater than the stream velocity.
Solution:
Let l be the distance covered by boat A along the river as well as by the boat B across the river. Let v_{0} be the stream velocity and v' the velocity of each boat with respect to water. Therefore time taken by the boat A in its journey
t_{A} = l/(v’ + v_{0}) + l/(v’ - v_{0})
t_{B }= l/√(v’^{2} - v_{0}^{2}) + l/√(v’^{2} - v_{0}^{2})
= 2l/√(v’^{2} - v_{0}^{2})
Hence t_{A}/t_{B} = v’/√(v’^{2} - v_{0}^{2})
= η/√(η^{2} - 1) (where η = v’/v)
On substitution t_{A}/t_{B} = 1.8
9. A boat moves relative to water with a velocity which is n = 2.0 times less than the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting?
Solution:
Let v_{0} be the stream velocity and v’ the velocity of boat with respect to water.
v_{0}/v’ = η = 2>0, some drifting of boat is inevitable.
Let
t = d/v’ sin θ (where d is the width of the river)
In this time interval, the drifting of the boat
x = (v’ cos θ + v_{0})t
= (v’ cos θ + v_{0}) d/v’ sin θ
= (cot θ + η cosec θ)d
For x_{min}(minimum drifting)
d/dθ (cot θ + η cosec θ) = 0, which yields
cos θ = -1/η
= -1/2
θ = 120^{0}
10. Two bodies were thrown simultaneously from the same point: one, straight up, and the other, at an angle of θ = 60° to the horizontal. The initial velocity of each body is equal to v_{o }= 25 m/s. Neglecting the air drag, find the distance between the bodies t = 1.70 s later.
Solution:
The solution of this problem becomes simple in the frame attached with one of the bodies. Let the body thrown straight up be 1 and the other body be 2, then for the body 1 in the frame of 2 from the kinematic equation for constant acceleration:
So,
But
So from the properties of triangle
v_{0(12) }= √(v_{0}^{2} + v_{0}^{2} - 2v_{0}v_{0} cos (π/2 - θ_{0})
Hence the sought distance
= 22 m
11. Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located at one point and moved with velocities v_{1} = 3.0 m/s and v_{2 }= 4.0 m/s horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.
Solution:
Let the velocities of the particles (say
Or -v_{1}v_{2} + g^{2}t^{2} = 0
Hence t = √(v_{1}v_{2})/g
12. Three points are located at the vertices of an equilateral triangle whose side equals a. They all start moving simultaneously with velocity v constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?
Solution:
From the symmetry of the problem all the three points are always located at the vertices of equilateral triangles of varying side length and finally meet at the centroid of the initial equilateral triangle whose side length is a, in the sought time interval (say t).
Let us consider an arbitrary equilateral triangle of edge length l (say). Then the rate by which 1 approaches 2, 2 approaches 3, and 3 approaches 1, becomes
-dl/dt = v - v cos (2π/3)
On integrating
a = (3/2) vt
So t = 2a/3v
13. Point A moves uniformly with velocity v so that the vector v is continually "aimed" at point B which in its turn moves rectilinearly and uniformly with velocity u< v. At the initial moment of time v ⊥ u and the points are separated by a distance l. How soon will the points converge?
Solution:
Let us locate the points A and B at an arbitrary instant of time (Fig.).
If A and B are separated by the distance s at this moment, then the points converge or point A approaches B with velocity -ds/dt = v - u cos α where angle α varies with time.
On integrating
Where T is the sought time.
As both A and B cover the same distance in x direction during the sought time interval, so the other condition which is required, can be obtained by the equation
So uT =
Solving (1) and (2), we get T = ul/(v^{2} - u^{2})
One can see that if u = v, or u < v, point A cannot catch B.
14. A train of length 1 = 350 m starts moving rectilinearly with constant acceleration w = 3.0.10-2 m/s^{2}; t = 30 s after the start the locomotive headlight is switched on (event 1) , and T = 60 s after that event the tail signal light is switched on (event 2) . Find the distance between these events in the reference frames fixed to the train and to the Earth. How and at what constant velocity V relative to the Earth must a certain reference frame K move for the two events to occur in it at the same point?
Solution:
In the reference frame fixed to the train, the distance between the two events is obviously equal to l. Suppose the train starts moving at time t = 0 in the positive x direction and take the origin (x = 0) at the head coordinate of first event in the earth’s frame is light o f the train at t = 0. Then the coordinate of first event in the earth’s frame is
x_{1} = ½ wt^{2}
And similarly the coordinate of the second event is
x_{2} = ½ w(t+T)^{2} - l
The distance between the two events is obviously
x_{1} - x_{2} = l - wT(t + T/2)
= 0.242 km
In the reference frame fixed frame on the earth.
For the two events to occur at the same point in the reference frame K, moving with constant velocity V relative to the earth, the distance travelled by the frame in the time interval T must be equal to the above distance.
Thus VT = l - wT(t + T/2)
So V = 1/T - w(t + T/2)
= 4.03 m/s
The frame K must clearly be moving in a direction opposite to the train so that if (for example) the origin of the frame coincides with the point x_{1} on the earth at time t, it_{coincides with the point x2 at time t + T.}
15. An elevator car whose floor-to-ceiling distance is equal to 2.7 m starts ascending with constant acceleration 1.2 m/s^{2}; 2.0 s after the start a bolt begins falling from the ceiling of the car. Find:
(a) the bolt's free fall time;
Solution:
(a) One good way to solve the problem is to work in the elevator’s frame having the observer at its bottom (Fig.).
Let us denote the separation between floor and ceiling by h = 2.7 m and the acceleration of the elevator by w = 1.2m/s^{2}
From the kinematical formula
y = y_{0} + v_{0y}t + ½ w_{y}t^{2} ..(1)
Here y = 0, y_{0} = +h, v_{0y} = 0
And w_{y} = bolt_{(y) }- w_{ele(y)}
= -g - w
= -(g + w)
So, 0 = h + ½ (-(g + w))t^{2}
Or t = √(2h/(g + w))
= 0.7 s
(b) At the moment the bolt loses contact with the elevator, it has already acquired the velocity equal to elevator, given by
v_{0} = (1.2)2
= 2.4 m/s
In the reference frame attached with the elevator shaft (ground) and pointing the y axis upward, we have for the displacement of the bolt,
∆y = v_{0y}t + ½ w_{y}t^{2}
= v_{0}t + ½ (-g)t^{2}
Or ∆y = (2.4)(0.7) + ½ (-9.8)(0.7)^{2}
= -0.7 m
Hence the bolt comes down or displaces downward relative to the point, when it loses contact with the elevator by the amount 0.7 m (Fig.). Obviously the total distance covered by the bolt during its free fall time.
s = |∆y| + 2(v_{0}^{2}/2g)
= 0.7 m + 2.4^{2}/9.8 m
= 1.3 m
16. Two particles, 1 and 2, move with constant velocities v_{1} and v_{2} along two mutually perpendicular straight lines toward the intersection point O. At the moment t = 0 the particles were located at the distances l_{1} and l_{2} from the point O. How soon will the distance between the particles become the smallest? What is it equal to?
Solution:
Let the particle 1 and 2 be at points B and A at t = 0 at the distances l_{1} and l_{2} from intersection point O . Let us fix the inertial frame with the particle 2. Now the particle 1 moves in relative to this reference frame with a relative velocity
From fig(b), v_{12} = √(v_{1}^{2} + v_{2}^{2}) and tan θ = v_{1}/v_{2}
The shortest distance
AP = AM sin θ
= (OA - OM)sin θ
= (l_{2} - l_{1} cot θ)sin θ
Or AP = (l_{2} - l_{1} v_{2}/v_{1})v_{1}/ √(v_{1}^{2} + v_{2}^{2})
= v_{1}l_{2} - v_{2}l_{1}/ √(v_{1}^{2} + v_{2}^{2}) (using 1)
The sought time can be obtained directly from the condition that (l_{1} - v_{1}t)^{2} + (l_{2} - v_{2}t)^{2} is minimum. This gives t = (l_{1}v_{1} + l_{2}v_{2})/(v_{1}^{2} + v_{2}^{2})
17. From point A located on a highway (Fig. 1.2) one has to get by car as soon as possible to point B located in the field at a distance l from the highway. It is known that the car moves in the field η times slower than on the highway. At what distance from point D one must turn off the highway?
Solution:
Let the car turn off the highway at a distance x from the point D. So, CD = x, and if the speed of the car in the field is v, then the time taken by the car to cover the distance AC = AD - x on the highway
t_{1} = (AD - x)/ηv ..(1)
The time taken to travel the distance CB in the field
t_{2} = √(l^{2} + x^{2})/v ..(2)
So the total time elapsed to move the car from point A to B
t = t_{1} + t_{2}
= (AD - x)/ηv + √(l^{2} + x^{2})/v
For t to be minimum
dt/dx = 0 or (1/v)(-1/η + x/√(l^{2} + x^{2})) = 0
Or η^{2}x^{2} = l^{2 }+ x^{2} or x = l/√(η^{2} - 1)
18. A point travels along the x axis with a velocity whose projection v_{x} is presented as a function of time by the plot in Fig. 1.3.
Assuming the coordinate of the point x = 0 at the moment t = 0, draw the approximate time dependence plots for the acceleration w_{x}, the x coordinate, and the distance covered s.
Solution:
To plot x(t), s(t) and w_{ x }(t) let us partition the given plot v_{x}(t) into five segments (for detailed analysis) as shown in the figure.
For the part oa: w_{x} = 1 and v_{x} = t = v
Thus ∆x_{1}(t) = ∫v_{x} dt = ∫_{0}^{t}dt = t^{2}/2 = s_{1}(t)
Putting t = 1, we get, ∆x_{1} = s = ½ unit
For the part ab:
w_{x} = 0 and v_{x} = v = constant = 1
Thus ∆x_{2} (t) = ∫v_{x} dt = ∫_{1}^{t} dt = (t - 1) = s_{2}(t)
Putting t = 3, ∆x_{2} = s_{2} = 2 unit
For part b_{4}: w_{x} = 1 and v_{x} = 1 - (t - 3) = 4 - t = v
Thus ∆x_{3}(t) = ∫_{3}^{t} (4 - t)dt = 4t - t^{2}/2 - 15/2 = s_{3}(t)
Putting t = 4, ∆x_{3} = x_{3 }= 1/2 unit
For part 4d: v_{x}= -1 and v_{x} = -(1-4) = 4 -1
So v = |v_{x}| = t - 4 for t> 4
Thus ∆x_{4}(t) = ∫_{4}^{t} (1 - t)dt = 4t - t^{2}/2 8= s_{3}(t)
Putting t = 6, ∆x_{4} = -1
Similarly s_{4}(t) = ∫|v_{x}| dt = ∫_{4}^{t}(t - 4)dt = t^{2}/2 - 4t + 8
Putting t = 6, s_{4} = 2 unit.
For part d7: w_{x} = 2 and v_{x} = -2 + 2(t - 6) = 2(t -7)
v = |v_{x}| = 2(7-t) for t ← 7
Now ∆x(t) = ∫_{t}^{6}2(t - 7) dt = t^{2} -14t + 48
Putting t = 4, ∆x_{5} = -1
Similarly s_{5}(t) = ∫_{t}^{6}2(7 - t)dt = 14t - t^{2} - 48
Putting t = 7, s_{5} = 1
On the basis of these obtained expressions w_{x}(t), x(t) and s(t) plots can be easily plotted as shown in the figure of answer sheet.
19. A point traversed half a circle of radius R = 160 cm during time interval T = 10.0 s. Calculate the following quantities averaged over that time:
if the point moved with constant tangent acceleration.
Solution:
(a) Mean velocity <v> = total distance covered/time elapsed
= s/t
= πR/T
= 50 cm/s
(b) Modulus of mean velocity vector =
= 2R/T = 32 cm/s (2)
(c) Let the point moves from i to f along the half circle (fig) and v_{0} and v be the speed at the points respectively.
We have dv/dt = w_{t}
Or v = v_{0} + w_{t}t (as w_{t} is constant, according to the problem)
20. A radius vector of a particle varies with time t as r = at (1 - αt), where a is a constant vector and α is a positive factor.
Find:
Solution:
(a) we have
So,
=
And
=
(b) from the equation
So the sought time ∆t = 1/α
As
Simplifying, we get s = a/2α
21. At the moment t = 0 a particle leaves the origin and moves in the positive direction of the x axis. Its velocity varies with time as v = v_{0} (1 - t/T), where v_{0} is the initial velocity vector whose
modulus equals v_{o} = 10.0 cm/s; T = 5.0 s. Find:
Solution:
As the particle leaves the origin at t = 0
So ∆x = x = ∫v_{x} dt
As
Where vector v_{0} is directed towards the +ve x axis
So v_{x} = v_{0}(1 - t/T)
From (1) and (2)
x = ∫_{0}^{t}v_{0}(1 - t/T)dt = v_{0}t(1-t/2T)
Hence x coordinate of the particle at t = 6s
x = 10×6(1-6/2×5)
= 24 cm
=0.24 m
Similarly at t = 10 s
x = 10×10(1-10/2×5)
= 0
And at t = 20 s
x = 10 ×20(1-20/2×5)
= -200 cm
= -2 m
b) At the moments the particle is at a distance of 10 cm from the origin, x = ±10 cm
Putting x = +10 in eq (3)
10 = 10t(1 - t/10) or t^{2} - 10t + 10 = 0
So t = 10±√(100 - 40)/2
= 5±√15 s
Now putting x = -10 in eqn (3)
-10 = 10(1-t/10)
On solving, t = 5±√35 s
As t cannot be negative, so
t = (5+√35) s
As t cannot be negative, so
t = 5 + √35 s
Hence the particle is at a distance of 10 cm from the origin at three moments of time:
t = 5±√15 s, 5+√35 s
On integrating and simplifying, we get s = 34 cm
On the basis of eqn (3) and (4), x(t) and s(t) plots can be drawn as shown in the answer sheet.
22. The velocity of a particle moving in the positive direction of the x axis varies as v = α√x, where α is a positive constant.
Assuming that at the moment t = 0 the particle was located at the point x = 0, find:
Solution:
(a) As particle is in unidirectional motion it is directed along the x axis at t = 0, x = 0 axis all the time.
So ∆x = x = s, and dv/dt = w
therefore , v = α√x = α√s
Or w = dv/dt
= (α/2√s)(ds/dt)
= α/2√s
= αv/2√s
= α^{2}/2
As w = dv/dt
= α^{2}/2
On integrating
Or v = α^{2}t/2
(b) Let s be the time to covers first s m of the path. From the eq.
s = ∫v dt
s = ∫_{0}^{t} α^{2}/2 dt
= (α^{2}/2) (t^{2}/2) using (2)
Or t = 2√s/α
The mean velocity of the particle
v = ∫v(t)dt/∫dt
= α√s/2
23. A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as w = a√v, where a is a positive constant. At the initial moment the velocity of the point is equal to v_{0}. What distance will it traverse before it stops? What time will it take to cover that distance?
Solution:
According to the problem
-v dv/ds = a√v (as v decreases with time)
Or
On integrating we get s = 2v_{0}^{3/2}/3a
Again according to the problem -dv/dt = a√v
Or -dv/√v = a dt
Or,
Thus t = 2√v_{0}/a
24. A radius vector of a point A relative to the origin varies with time t as r = ati - bt^{2}j, where a and b are positive constants, and i and j are the unit vectors of the x and y axes. Find:
Solution:
(a) As
x = at, y = -bt^{2}
And therefore y = -bx^{2}/a^{2}
which is Eq. of a parabola, whose graph is shown in the fig.
(b) As
So v = √(a^{2} + (2bt)^{2}) = √(a^{2} + 4b^{2}t^{2})
Diff. Eq.(1) w.r.t.time, we get
So |w| = 2b
(c) cos α =
cos α = 2bt/√(a^{2}+4b^{2}t^{2})
So tan α = a/2bt
Or α = tan^{-1} (a/2bt)
(d) the mean velocity vector
=
Hence
= √(a^{2} + b^{2}t^{2})
25. A point moves in the plane xy according to the law x = at, y = at (1-αt), where a and α are positive constants, and t is time. Find:
Solution:
(a) We have x = at and y = αt (1- αt) ..(1)
Hence, y (x) becomes,
y = (ax/a)(1 - αx/a)
= x - αx^{2}/a (parabola)
(b) Differentiating Eq. (1) we get
v_{x} = a and v_{y} = a(1 - 2αt)
So v = √(v_{x}^{2} + v_{y}^{2})
= a√(1 + (1 - 2αt)^{2})
Diff. Eq. (2) w.r.t.time
w_{x} = 0 and w_{y} = -2aα
So w = √(w_{x}^{2} + w_{y}^{2})
= 2aα
(c) From Eqs. (2) and (3)
We have
And
So cos π/4 = 1/√2 =
= -a(1 - 2αt_{0})2aα/a√(1 + (1 - 2αt_{0})^{2}2aα
On simplifying
1 - 2αt_{0} = ±1
As, t ≠ 0, t_{0} = 1/α
26. A point moves in the plane xy according to the law x = a sin ωt, y = a (1 - cos ωt), where a and ω are positive constants. Find:
Solution:
Differentiating motion law : x = a sin ωt , y = a (1 - cos ωt ), with respect to time, v_{x}= a ω cos ωt, v_{y} = a ω sin ωt
So
And v = aω = constant ..(2)
Differentiating Eq. (1) with respect to time
(a) the distance s traversed by the point during the time T is given by
s = ∫_{0}^{T}v dt = ∫_{0}^{T}aω dt = aωT (using 2)
(b) Taking inner product of vector v and vector w
We get
So
Thus
So the angle between velocity vector and acceleration vector equals π/2.
27. A particle moves in the plane xy with constant acceleration w directed along the negative y axis. The equation of motion of the particle has the form y = ax - bx^{2}, where a and b are positive constants. Find the velocity of the particle at the origin of coordinates.
Solution:
According to the problem
So, w_{x} = dv_{x}/dt = 0 and w_{y} = dv_{y}/dt = -w ..(1)
Differentiating Eq. of trajectory, y = ax - bx^{2}, with respect to time
dy/dx = a dx/dt - 2bx dx/dt ..(2)
So, dy/dx _{at (x = 0)} = a dx/dt_{at x = 0}
Differentiating w.r.t.time
d^{2}y/dx^{2} = ad^{2}x/dt^{2} - 2b(dx/dt)^{2} - 2bx d^{2}x/dt^{2}
Or -w = a(0) - 2b(dx/dt)^{2} - 2bx(0) (using 1)
Or dx/dt = √(w/2b) (using (1)) (3)
Using (3) in (2) dy/dt_{x=0} = a √(w/2b) ..(4)
Hence the velocity of the particle at the origin
v = √((dx/dt)^{2}_{x=0} + (dy/dt)^{2}_{x=0} )
= √((w/2b) + (a^{2}w/2b))
Hence v = √((w/2b)(1+a^{2}))
28. A small body is thrown at an angle to the horizontal with the initial velocity v_{o}. Neglecting the air drag, find:
Solution:
As the body is under gravity of constant acceleration g, its velocity vector and displacement vectors are
But we have v = v_{0} at t = 0 and also at t = T (also from energy conservation)
Hence using this property in Eq. 5
29. A body is thrown from the surface of the Earth at an angle α to the horizontal with the initial velocity v_{0}. Assuming the air drag to be negligible, find:
Solution:
The body thrown in air with velocity v_{0} at an angle α from the horizontal lands at point P on the Earth's surface at same horizontal level (Fig.). The point of projection is taken as origin, so, Δx = x and Δy = y
(a) from the eq. ∆y = v_{0y}t + ½ w_{y}t^{2}
0 = v_{0} sin αT - ½ gT^{2}
At T not equal 0, so time of motion T = 2v_{0} sin α/g
(b) At the max. Height of ascent, v_{y} = 0
So from the Eq. v_{y}^{2} = v_{0y}^{2} + 2w_{y}∆y
0 = (v_{0}sin α)^{2} - 2gH
Hence maximum height H = v_{0}^{2} sin^{2} α/2g
During the time of motion the net horizontal displacement or horizontal range, will be obtained by the equation
∆x = v_{0x} t + ½ w_{x}T^{2}
Or R = v_{0} cos αT - ½ (0)T^{2}
= v_{0}cos αT
= v_{0}^{2}sin 2α/g
When R = H
v_{0}^{2}sin^{2}α/g = v_{0}^{2}sin^{2}α/2g
Or tan α = 4
So α = tan^{-1} 4
(c) For the body, x(t) and y(t) are x = v_{0}cos αt
And y = v_{0} sin αt - ½ gt^{2}
Hence putting the value of t from (1) and (2) we get,
y = v_{0} sin α(x/v_{0} cos α) - ½ g(x/v_{0}cos α)^{2} = x tan α - gx^{2}/2v_{0}^{2}cos^{2}α
Which is the sought equation of trajectory i.e y(x)
(d) As the body thrown in air follows a curve, it has some normal acceleration at all the moments of time during it’s motion in air. At the initial point (x = 0, y = 0), from the equation
w_{n} = v^{2}/R (where R is the radius of curvature)
g cos α = v_{0}^{2}/R_{0} (see fig)
Or R_{0} = v_{0}^{2}/g cos α
At the peak point v_{y} = 0, v = v_{x} = v_{0} cos α and the tangential acceleration is zero.
Now from the Eq. w_{n} = v^{2}/R
g = v_{0}^{2} cos^{2}α/R, or R = v_{0}^{2} cos^{2}α/g
Note: We may use the formula of curvature radius of a trajectory y(x)_{n} to solve part (d)
R = [1 + (dy/dx)^{2}]^{3/2}/|d^{2}y/dx^{2}|