Motion in a Straight Line

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 :

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 :-

A car accelerates from rest at a constant rate $$\alpha$$ for some time after which it decelerates at a constant rate $$\beta$$ to come to rest. If the total time elapsed is t seconds, the total distance travelled 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

An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be

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

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) :

In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed '$$\upsilon$$' more than that of car B. Both the cars start from rest and travel with constant acceleration a₁ and a₂ respectively. Then '$$\upsilon$$' is equal to :