(D) Let the velocity of body $A$ be $v_A = u$ (constant).Let the velocity of body $B$ be $v_B = at$ (starting from rest with acceleration $a$).Their v

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(D) Let the velocity of body $A$ be $v_A = u$ (constant).Let the velocity of body $B$ be $v_B = at$ (starting from rest with acceleration $a$).Their velocities become equal when $v_A = v_B$,which implies $u = at$. Thus,the time taken is $t = \frac{u}{a}$.The distance covered by body $A$ in time $t$ is $s_A = u \cdot t = u \left( \frac{u}{a} \right) = \frac{u^2}{a}$.The distance covered by body $B$ in time $t$ is $s_B = \frac{1}{2} a t^2 = \frac{1}{2} a \left( \frac{u}{a} \right)^2 = \frac{u^2}{2a}$.The distance between them is $d = s_A - s_B = \frac{u^2}{a} - \frac{u^2}{2a} = \frac{u^2}{2a}$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

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Two bodies $A$ and $B$ move in the same straight line starting from the same position. Body $A$ moves with a constant velocity $u$ and body $B$ moves with a constant acceleration $a$ starting from rest. When their velocities become equal,the distance between them is:

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A
$\frac{u}{2a}$
B
$\frac{u^2}{4a}$
C
$\frac{u^2}{a}$
D
$\frac{u^2}{2a}$

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Motion in Straight Line

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Uniformly Accelerated Motion

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$A$ body of mass $10 \, kg$ is moving with a constant velocity of $10 \, m/s$. When a constant force acts for $4 \, s$ on it,it moves with a velocity of $2 \, m/s$ in the opposite direction. The acceleration produced in it is .......... $m/s^2$.

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$A$ body starting from rest moves with an acceleration of $\frac{5}{4} \,ms^{-2}$. The distance travelled by the body in the third second is:

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$A$ car is moving along a straight road with uniform acceleration. It passes through two points $P$ and $Q$ separated by a distance $s$ with velocities $30\; km/h$ and $40\; km/h$ respectively. Find the velocity of the car midway between $P$ and $Q$.

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$A$ 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 $\frac{f}{2}$ to come to rest. If the total distance traversed is $15S$,then:

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Two bodies $A$ and $B$ move in the same straight line starting from the same position. Body $A$ moves with a constant velocity $u$ and body $B$ moves with a constant acceleration $a$ starting from rest. When their velocities become equal,the distance between them is:

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$A$ body of mass $10 \, kg$ is moving with a constant velocity of $10 \, m/s$. When a constant force acts for $4 \, s$ on it,it moves with a velocity of $2 \, m/s$ in the opposite direction. The acceleration produced in it is .......... $m/s^2$.

$A$ body starting from rest moves with an acceleration of $\frac{5}{4} \,ms^{-2}$. The distance travelled by the body in the third second is:

$A$ car is moving along a straight road with uniform acceleration. It passes through two points $P$ and $Q$ separated by a distance $s$ with velocities $30\; km/h$ and $40\; km/h$ respectively. Find the velocity of the car midway between $P$ and $Q$.

$A$ 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 $\frac{f}{2}$ to come to rest. If the total distance traversed is $15S$,then:

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