Motion in a Straight Line

A ball is released from a height h. If $t_{1}$ and $t_{2}$ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between $t_{1}$ and $t_{2}$.

A car starting from rest accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate ${f \over 2}$ to come to rest. If the total distance traversed is 15 S, then

The position vector of a particle changes with time according to the relation $\overrightarrow r (t) = 15{t^2}\widehat i + (4 - 20{t^2})\widehat j$ What is the magnitude of the acceleration at t = 1 ?

Given below are two statements Statement I : Area under velocity- time graph gives the distance travelled by the body in a given time. Statement II : Area under acceleration- time graph is equal to the change in velocity- in the given time. In the light of given statements, choose the correct answer from the options given below.

A vehicle travels $4 \mathrm{~km}$ with speed of $3 \mathrm{~km} / \mathrm{h}$ and another $4 \mathrm{~km}$ with speed of $5 \mathrm{~km} / \mathrm{h}$, then its average speed is

A car travels a distance of '$x$' with speed $v_1$ and then same distance '$x$' with speed $v_2$ in the same direction. The average speed of the car is :

The distance travelled by an object in time $t$ is given by $s=(2.5) t^{2}$. The instantaneous speed of the object at $\mathrm{t}=5 \mathrm{~s}$ will be:

A particle moving in a straight line covers half the distance with speed $6 \mathrm{~m} / \mathrm{s}$. The other half is covered in two equal time intervals with speeds $9 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$ respectively. The average speed of the particle during the motion is :

An automobile, travelling at $40\,$ km/h, can be stopped at a distance of $40\,$ m by applying brakes. If the same automobile is travelling at $80\,$ km/h, the minimum stopping distance, in metres, is (assume no skidding) :

A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching ${h \over 3}$ in both the directions.

A train starting from rest first accelerates uniformly up to a speed of $80 \mathrm{~km} / \mathrm{h}$ for time $t$, then it moves with a constant speed for time $3 t$. The average speed of the train for this duration of journey will be (in $\mathrm{km} / \mathrm{h}$) :

The co-ordinates of a moving particle at any time 't' are given by x = $\alpha$t³ and y = βt³. The speed to the particle at time 't' is given by

The instantaneous velocity of a particle moving in a straight line is given as $V = \alpha t + \beta {t^2}$, where $\alpha$ and $\beta$ are constants. The distance travelled by the particle between 1s and 2s is :

If $\mathrm{t}=\sqrt{x}+4$, then $\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)_{\mathrm{t}=4}$ is :

The distance travelled by a particle is related to time t as $x=4\mathrm{t}^2$. The velocity of the particle at t=5s is :-

The position of a particle related to time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$. The magnitude of velocity of the particle at $t=2 s$ will be :

A person travels $x$ distance with velocity $v_{1}$ and then $x$ distance with velocity $v_{2}$ in the same direction. The average velocity of the person is $\mathrm{v}$, then the relation between $v, v_{1}$ and $v_{2}$ will be.

Speeds of two identical cars are $u$ and $4$$u$ at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is :

A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is