A body $$A$$ begins to move with initial velocity $$2\,m/\sec $$ and continues to move at a constant acceleration $$a.$$ $$\Delta t = $$ 10 seconds after the body $$A$$ begins to move a body $$B$$ departs from the same point with an initial velocity $$12\,m/\sec $$ and moves with the same acceleration $$a.$$ What is the maximum acceleration $$a$$ at which the body $$B$$ can overtake $$A$$ ?
A.
$$1\,m/{s^2}$$
B.
$$2\,m/{s^2}$$
C.
$$\frac{1}{2}\,m/{s^2}$$
D.
$$3\,m/{s^2}$$
Answer :
$$1\,m/{s^2}$$
Solution :
$$a = \frac{{{{v''}_0} - {{v'}_0}}}{{\Delta t}}$$
On substituting
$${{v'}_0} = 2\,m/s,{{v''}_0} = 12\,m/s$$ and $$\Delta t = 10\,\sec ,$$
we get, $$a = 1\,m/{s^2}.$$
Releted MCQ Question on Basic Physics >> Kinematics
Releted Question 1
A river is flowing from west to east at a speed of $$5$$ metres per minute. A man on the south bank of the river, capable of swimming at $$10$$ metres per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction-
A boat which has a speed of $$5 km/hr$$ in still water crosses a river of width $$1 \,km$$ along the shortest possible path in $$15 \,minutes.$$ The velocity of the river water in $$km/hr$$ is-
In $$1.0\,s,$$ a particle goes from point $$A$$ to point $$B,$$ moving in a semicircle of radius $$1.0 \,m$$ (see Figure). The magnitude of the average velocity-
A ball is dropped vertically from a height $$d$$ above the ground. It hits the ground and bounces up vertically to a height $$\frac{d}{2}.$$ Neglecting subsequent motion and air resistance, its velocity $$v$$ varies with the height $$h$$ above the ground as-