AIPMT-PRELIMS Physics Motion in a Straight Line Class 11 Questions

A particle has initial velocity \(-2\hat{i} + 3\hat{j}\) and acceleration \(0.3\hat{i} + 0.2\hat{j}\). The magnitude of velocity after 10 seconds will be

The motion of a particle along a straight line is described by equation: x = 8 + 12t - t³ where x is in metre and t in second. The retardation of the particle when its velocity becomes zero, is

A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m/s², the velocity with which it hits the ground is

A particle moves a distance x in time t according to equation \( x = (t + 5)^{-1} \). The acceleration of particle is proportional to

A particle has initial velocity (3î + 4ĵ) and has acceleration (0.4î + 0.3ĵ). Its speed after 10 s is

A ball is dropped from a height h at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? (Take g = 10 m/s²)

A bus is moving with a speed of 10 ms⁻¹ on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase the bus?

A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is S₁ and that covered in the first 20 seconds is S₂, then:

The distance travelled by a particle starting from rest and moving with an acceleration 4/3 m/s², in the third second is -

A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms⁻¹ to 20 ms⁻¹ while passing through a distance 135 m in t second. The value of t is -

A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point :

The position x of a particle with respect to time t along x-axis is given by x = 9t² - t³ where x is in metres and t in second. What will be the position of this particle when it achieves maximum speed along the +ve x direction?

A car moves from X to Y with a uniform speed vᵤ and returns to Y with a uniform speed v_d. The average speed for this round trip is

A particle moving along x-axis has acceleration f, at time t, given by f = f₀ [¹⁻ (t/T)], where f₀ and T are constants. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle’s velocity (vₓ) is

A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by x = 40 + 12t - t³. How long would the particle travel before coming to rest?

A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The average velocity and average speed for each circular lap respectively is:

Two bodies, A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is:

The displacement x of a particle varies with time t as x = ae^{-αt} + be^{βt}, where a, b, α and β are positive constants. The velocity of the particle will:

Two boys are standing at the ends A and B of a ground, where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity v1. The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is :

A ball is thrown vertically upward. It has a speed of 10 m/s when it has reached one half of its maximum height. How high does the ball rise? (Taking g = 10 m/s²)

If a ball is thrown vertically upwards with speed u, the distance covered during the last ‘t’ seconds of its ascent is :

A man throws ball with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time ? (Given g = 9.8 m/s²)

A particle is thrown vertically upward. Its velocity at half of the height is 10 m/s. Then the maximum height attained by it: (g = 10 m/s²)

Motion of a particle is given by equation s = (3t³ + 7t² + 14t + 8)m. The value of acceleration of the particle at t = 1 sec. is :