# Acceleration

## Trending Questions

**Q.**The displacement is given by x=2t2+t+5 in m, the acceleration at t=5s will be

- 8 m/s2
- 12 m/s2
- 15 m/s2
- 4 m/s2

**Q.**The displacement x of a particle along a straight line at time t is given by x=a0–a1t+a2t2. The acceleration of the particle is:

- a0
- a1
- 2a2
- a2

**Q.**

Can the velocity be 0 but acceleration be non zero?

Yes

No

- In space
- More information needed

**Q.**A train accelerates from 36 km/h to 54 km/h in 5 s then average acceleration of the train is

- 1 m/s2
- 2 m/s2
- 0.5 m/s2
- 1.5 m/s2

**Q.**

This section contains 1 Assertion-Reason type question, which has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

इस खण्ड में 1 कथन-कारण प्रकार का प्रश्न है, जिसमें 4 विकल्प (a), (b), (c) तथा (d) दिये गये हैं, जिनमें से केवल एक सही है।

A : एक वस्तु का तब भी त्वरण हो सकता है, जब इसकी चाल एकसमान हो।

R : If velocity is constant, then there is no change in direction of motion.

R : यदि वेग नियत है, तब गति की दिशा में कोई परिवर्तन नहीं होता है।

- Both (A) and (R) are true and (R) is the correct explanation of (A)

(A) तथा (R) दोनों सही हैं तथा (R), (A) का सही स्पष्टीकरण है - Both (A) and (R) are true but (R) is not the correct explanation of (A)

(A) तथा (R) दोनों सही हैं लेकिन (R), (A) का सही स्पष्टीकरण नहीं है - (A) is true but (R) is false

(A) सही है लेकिन (R) गलत है - (A) is false but (R) is true

(A) गलत है लेकिन (R) सही है

**Q.**The motion of a body is given by the equation dvdt=6−3v, where v is speed in m/s and t is time in seconds. If the body is at rest at t=0, then

- the terminal speed is 2 m/s.
- the speed varies with the time as v=2(1−e−3t) m/s.
- the speed is 0.1 m/s when the acceleration is half the initial value.
- the magnitude of the initial acceleration is 6 m/s2.

**Q.**Velocity of the particle in a rectilinear motion is given as v=3t4−5t in m/s. Find

(i) Acceleration at 2 sec.

(ii) Average acceleration in 2 sec.

- (i) 71 m/s2

(ii) 17 m/s2 - (i) 81 m/s2

(ii) 18 m/s2 - (i) 91 m/s2

(ii) 19 m/s2 - (i) 61 m/s2

(ii) 16 m/s2

**Q.**The relation between time t and distance x is t=αx2+βx where α and β are constants. The retardation is

- 2 αv3
- 2 βv3
- 2 αβv3
- 2 β2v3

**Q.**Instantaneous velocity of an object (starting from origin) varies with time as v=α−βt2(where α and β are positive constants). If x and a are the position and acceleration of the object as a function of time. Then,

- x=αt−13βt3
- x=αt−12βt2
- α=−βt
- Maximum positive displacement from origin =2α3√αβ

**Q.**A particle moving eastwards with 5 ms−1. In 10 s the velocity changes to 5 ms−1 northwards. The average acceleration in this time is

- 1√2 m/s2 North-East.
- 12 m/s2 North-West.
- 1√2 m/s2 North-West.
- Zero.

**Q.**A particle starts from x=0 and moves along a straight line such that its velocity v depends on position x as v=x+4.

Acceleration of the particle at t=0 is

- e4 m/s2
- 4e4 m/s2
- 4 m/s2
- 0

**Q.**Acceleration of the particle in a rectilinear motion is given as a=4t+3 in (m/s2). The velocity and displacement as a function of time t respectively are

(Given that at t=0, u=3 m/s, x=0 m)

- (t2+3), (t33+3t22+3t)
- (t2+3t), (t33+3t22)
- (2t2+3t+3), (2t33+3t22+3t)
- (2t2+3t), (2t33+3t22)

**Q.**For a particle moving along x - axis, the acceleration a of the particle in terms of its x- coordinate x is given by a=−9x, where x is in meters and a is in m/s2. Take acceleration, velocity and displacement in positive x - direction as positive. The initial velocity of particle at x=0 is u=+6 m/s. The velocity of particle at x=2 m will be

- +6√2 m/s
- −6√2 m/s
- 72 m/s
- 0

**Q.**The displacement is given by x=2t2+t+5 in m, the acceleration at t=5s will be

- 8 m/s2
- 12 m/s2
- 15 m/s2
- 4 m/s2

**Q.**Match Column - I with Column -II

Column - IColumn - IIi.Velocity of a particle following (p) αequation x=u(t−8)+α(t−2)2at t=0ii.Acceleration of a particle moving according to equation(q)2αx=u(t−8)+α(t−2)2 at t=0iii.Velocity of the particle moving (r) uaccording to equation x=ut+5αt2ln(1+t2)+αt2at t=0iv.Acceleration of the particle moving (s) u−4αaccording to the equation x=ut+5αt2(1+t2)+αt2at t=0

- i→(s);ii→(q);iii→(r);iv→(q)
- i→(s);ii→(q);iii→(s);iv→(p)
- i→(r);ii→(p);iii→(s);iv→(q)
- i→(r);ii→(r);iii→(r);iv→(q)

**Q.**Acceleration of the particle in a rectilinear motion is given as a=2t−6 in (m/s2). The displacement as a function of time t respectively is

(Given, at t=0, u=2 m/s, x=2 m)

- 2t33−3t2+2t+2
- t33−3t2+2t
- t33−3t2+2t+2
- t33−3t2

**Q.**A particle moves in XY plane such that its position, velocity and acceleration are given by →r=x^i+y^j; →v=vx^i+vy^j; →a=ax^i+ay^j

which of the following condition is correct if the particles speed is reducing?

- xvx+yvx<0
- xvx+yvy+yvy>0
- axvx+ayvy<0
- axvx+ayvy>0

**Q.**The plot of velocity versus time graph for a particle moving along a straight line is shown.

The value of dot product of velocity and acceleration of particle at t=2 s is

- 1 m2/s3
- −1 m2/s3
- 2 m2/s3
- −2 m2/s3

**Q.**The displacement x of a particle along a straight line at time t is given by x=a0+a1t+a2t3. The acceleration a of the particle at time t=5 is

- 15a2
- 30a2
- a1+30a2
- 6a2

**Q.**The x coordinates of a particle at any time t are given by x=7t+4t2 where x is in m and t in s. The acceleration of the particle at 5 s is

- zero
- 8 m/s2
- 20 m/s2
- 40 m/s2

**Q.**A particle starting from rest undergoes acceleration given by a=|t–2| m/s2 where, t is time in second. Velocity of particle after 4 sec is

- 1 m/s
- 2 m/s
- 8 m/s
- 4 m/s

**Q.**Acceleration of the particle in a rectilinear motion is given as a=2t−2 in (m/s2). The final position of the particle as a function of time t is

(Given that at t=0, u=2 m/s, x=4 m)

- t33−t2+2t+4
- t33−t2+2t
- t33−t2+4
- t33−t2

**Q.**The position of the particle is given by x=2t3−4t2+5 in m. The acceleration of the particle at 5 sec is

- 45 m/s2
- 48 m/s2
- 40 m/s2
- 52 m/s2

**Q.**

A car is moving along a straight line. It's displacement (x) - time (t) graph is shown. Match the entries in column I with points on graph as in Column-II.

Column I

(p) x → negative, v → positive, a → positive (q) x → positive, v → negative, a → negative

(r) x → negative, v → negative, a → positive (s) x → positive, v → positive, a → negative.

Column II

(p - d) (q - b) (r - c) (s - a)

(p - b) (q - d) (r - c) (s - a)

(p - a) (q - c) (r - b) (s - d)

(p - c) (q - a) (r - d) (s - b)

**Q.**If the pulley is massless and moves with an upward acceleration ao, find the acceleration of block m1 w.r.t elevator.

- (m2−m1)(g−ao)(m1+m2)
- (m2−m1)(g+ao)(m1+m2)
- (m2−m1)gm1
- (m2−m1)aom1

**Q.**Acceleration of the particle in a rectilinear motion is given as a=4t+3 in (m/s2). The final position of the particle as a function of time t is

(Given, at t=0, u=3 m/s, x=2 m)

- 2t33+3t22+3t
- 2t33+3t22
- 2t33+3t22+3t+2
- 2t33+3t22−3t

**Q.**A car start from rest and acquire a velocity of 54 km/h in 2 sec. Find the acceleration of the car.

- 5 m/s2
- 7.5 m/s2
- 15 m/s2
- 7 m/s2

**Q.**The position x of a particle varies with time t as x=ae−αt+beβt, where a, b, α and β are positive constants. The velocity of the particle will

- go on decreasing with time.
- be independent of α and β .
- drop to zero when α=β.
- go on increasing with time.

**Q.**The velocity displacement curve for an object moving along a straight line is shown in the given figure. At which of the marked points, the object is speeding up?

- 1
- 2
- 1 and 3
- 1, 2 and 3

**Q.**The pulley arrangements shown in figure are identical. The mass of the rope being negligible. In case-I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case-II, the mass m is lifted by pulling the other end of the rope with constant downward force F=2mg, where g is acceleration due to gravity. The acceleration of mass m in case-I is

more than that in case II

- equal to that in case II
Zero