Motion in a straight line practice · Medium

A particle is thrown vertically upward with speed 45 m/s. Find the distance covered by the particle in the 5th second of its journey.

If $f=f_0\left[1-\frac{t}{T}\right]$, where $f=$ acceleration at time $t$ then find value of $v$ at $t=t$. when $\mathbf{t}=\mathbf{0}, \mathbf{u}=\mathbf{0}$

A particle starts from rest at the point $A(1,2,-3) {m}$ and reaches at the point $B(3,4,5) {m}$. The magnitude of displacement of man is

Which branch of physics deals with motion without considering its causes?

A ball is thrown vertically upwards with a speed of 20 m/s from the top of a tower 250 m height and another is thrown vertically downwards with the same speed simultaneously. The time difference between them $m$ reaching the ground is

A plane travelling north at $200 \mathrm{~m} / \mathrm{s}$ turns and then travels south at $200 \mathrm{~m} / \mathrm{s}$. Its change in velocity is:

A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of $254\mathrm{m}{\mathrm{s}^{ - 1}}$. The two stones will meet after

A body is projected up with a speed ‘u' and the time taken by it is T to reach the maximum height H. Pick out the correct statement

A particle under the action of a constant force moves from rest upto 20 seconds. If distance covered in first 10 seconds is $\mathrm{S}_1$ and that covered in next 10 seconds in $\mathrm{S}_2$, then

A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is

Three ships A, B and C are in motion. The motion of A as seen by B is with speed V towards the north-east. The motion of B as seen by C is with speed V towards the north-west. Then, as seen by A, C will be moving towards the

A body starts moving from rest with constant acceleration covers displacement $S_1$ in first $(p-1)$ seconds and $S_2$ in first $p$ seconds. The displacement $S_1+S_2$ will be made in time:

A drunkard takes a step of $1 \mathrm{~m}$ in 1 second. He takes 5 seconds steps forward and 3 seconds steps backwards, and so on. The time taken by him to fall in a pit $13 \mathrm{~m}$ away from the starts is

Stopping distance of a moving vehicle is directly proportional to

Assertion : A positive acceleration of a body can be associated with 'slowing down' of the body.

Reason : Acceleration is a vector quantity.

Assertion: If a particle is thrown upwards then distance travelled in last second of upward journey is independent of the velocity of projection.
Reason: In last second distance travelled is $4.9 \mathrm{~m}$. (take $g=9.8 \mathrm{~m} / \mathrm{s}^2$)

The distance $x$ (in $\mu {m}$ ) covered by a molecule starting from point ${A}$ at time ${t}=0$ and stopping at another point ${B}$ in given by the equation

$x=t^2\left(2+\frac{t}{3}\right)$

The distance between $A$ and $B$ (in $\mu {m}$ ) is closed to

A police jeep is chasing culprit going on motorcycle. The motorcycle crosses a turning at $20 \mathrm{~m} / \mathrm{s}$. The jeep follows at speed of $30 \mathrm{~m} / \mathrm{s}$, crossing the turning 6 second later than motor cycle. If they travel at constant speeds, how far from turning point jeep catches up the motorcycle?

A bird flies due south with velocity $30 {~m} / {s}$ for 10 ${s}$, it rests for $2 {~s}$ and then flies due north with velocity $32 {~m} / {s}$ for $8 {~s}$. The average velocity of the bird

(in ${m} / {s}$ ) for the whole trip is;

(Consider north direction along positive y axis.)

Assertion (A): Starting from rest with zero acceleration if acceleration of particle increases at a constant rate of $2 \mathrm{~ms}^{-3}$ then velocity should increase at constant rate of $1 \mathrm{~ms}^{-2}$.

Reason (R): For the given condition.

$\frac{d a}{d t} =2 \mathrm{~ms}^{-3}$

$a =2 t$

Assertion: For an object having uniformly accelerated motion, the position-time graph is parabolic in nature.

Reason: In a uniformly accelerated motion, the acceleration is constant.

Two balls are dropped from the top of a cliff at a time interval $\Delta t=2 \mathrm{~s}$. The first ball hits the ground, rebounds elastically (essentially reversing direction instantly without losing speed) and collides with the second ball at height $h=55 \mathrm{~m}$ above the ground. How high is the top of the cliff (in $\mathrm{m}$ )?

A particle moves along a straight line such that its displacement at any time $t$ given by

$s=\left(\frac{t^4}{2}-12 t^2\right) m$

The velocity when the acceleration becomes zero is

A body is moving with uniform acceleration covers 200 m in the first 2 s and 220 m in the next 4 s. find the velocity in $\mathrm{m}{\mathrm{s}^{ - 1}}$ after 7 s.

A train starts from rest and moves with a constant acceleration of $2.0 \mathrm{~m} / \mathrm{s}^2$ for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed.

A particle is constrained to move on a straight line path. It returns to the starting point after 10 sec. The total distance covered by the particle during this time is 30 m. Which of the following statement about the motion of the particle is false

Assertion : A body may be accelerated even when it is moving uniformly.

Reason : When direction of motion of the body is changing, the body must have acceleration.

If a body loses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest

The velocity of a body depends on time according to the equation $v=40+0.2 t^3$. The body is undergoing

A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10 m in $t\,s$, the distance travelled by the toy in the next $t\,s$ will be :