Uniform and Non Uniform
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A bullet is fired on a wall with velocity 100ms-1. If the bullet stops at a depth of 10cm inside the wall, find the retardation produced by the wall.
When a particles distance traveled is proportional to the squares of time, the particle travels with
Uniform acceleration
Uniform velocity
Increasing acceleration
Decreasing velocity
The distance traveled by car in seconds after applying the brakes is feet, where . The car travels a certain distance before coming to a complete halt.
The equation of motion of a car is , where is measured in hours and s in kilometres. When the distance travelled by a car is , the velocity of the car is
The following relation [in MKS units] gives the distance traversed s by an accelerated particle of mass . The particles velocity after seconds in the appropriate unit is
is the distance traveled by an item along a straight line in time , and the beginning velocity of the object is
- Maximum velocity of balloon will be √a0H0
- Maximum velocity of balloon will be 2√a0H0
- Velocity when it is at height H02 is √3a0H02
- Velocity when it is at height H02 is √a0H0
- 24 metres
- 12 metres
- 5 metres
- Zero
The equation of motion of a car is , where t is measured in hours and s is measured in kilometers. If the distance traveled by car is 15 km, find the velocity of the car
- The body comes to rest first at (3–√7)s and then at (3+√7)s
- The total displacement of the particle in travelling from the first zero of velocity to the second zero of velocity is zero
- The total displacement of the particle in travelling from the first zero of the velocity to the second zero of velocity is –74 cm
- The particle reverses it’s velocity at (3–√7)s and then at (3+√7)s and has a negative velocity for (3–√7)s<t<(3+√7)s
- Initial velocity of particle is zero
- Motion is non-uniformly accelerated
- Acceleration of particle at x=2 m is 12 m/s2
- Acceleration of particle at x=4 m is 1 m/s2
- Increases
- Decreases
- Stays constant
- First decreases and then increases
Then the magnitude of displacement (in meters) by the particle from time t=0 to t=t will be
- 2π2sinπt−2tπ
- −2π2sinπt+2tπ
- 2tπ
- None of these
- Sv+v2(1α+1β)
- Sv+vα+vβ
- (vα+vβ)
- Sv−v2(vα+vβ)
- The displacement (x)=∫vdvf(v)
- The acceleration may be constant
- The slope of acceleration versus velocity graph may be constant.
- (a) and (c) are correct.
The velocity of a particle varies with time at v(t)=e−2πt. Find the displacement of the particle till t=(12π) sec.
−(1)2πe
12πee
(e−1)2πe
1e
- zero
- 8 m/s2
- 20 m/s2
- 40 m/s2
- −9x m/s2
- −18x m/s2
- −9x2 m/s2
- None of these
x=2t3−6t2+12t+16
The acceleration of the body is zero at time t is equal to
- 1 s
- 2 s
- 3 s
- 0.5 s
- Maximum velocity of balloon will be √a0H0
- Maximum velocity of balloon will be 2√a0H0
- Velocity when it is at height H02 is √3a0H02
- Velocity when it is at height H02 is √a0H0
- increases linearly
- decreases linearly
- remains constant
- first increases and then decreases
- Increasing acceleration
- Decreasing acceleration
- Increasing retardation
- Decreasing retardation
(Assume SI units)
- v=r
- v=2r
- v=(n+3)r2
- v=(n+1)r2
- Curve 1 may represent acceleration against time graph
- Curve 2 may represent velocity against time graph
- Curve 2 may represent velocity against acceleration graph
- None of these
- The initial velocity of the particle is zero
- The initial velocity of the particle is 2.5 ms−1
- The acceleration of the particle is 2.5 ms−2
- The acceleration of the particle is 50 ms−2
- 1x
- 1x3
- −1x2
- −1x3
- The initial velocity of the particle is zero
- The initial velocity of the particle is 2.5 ms−1
- The acceleration of the particle is 2.5 ms−2
- The acceleration of the particle is 50 ms−2
Then the distance travelled (in meters) by the particle from time t=0 to t=t will be
- 2π2sinπt−2tπ
- −2π2sinπt+2tπ
- 2tπ
- None of these
- Increases
- Decreases
- Stays constant
- First decreases and then increases
- 2 m/s
- 5 m/s
- 3 m/s
- 4 m/s