$A$ vehicle moving with $15 \,km/hr$ comes to rest by covering $5 \,m$ distance by applying brakes. If the same vehicle moves at $45 \,km/hr$, then by applying brakes, it will come to rest by covering a distance of: (in $\,m$)

$A$ car starts at time $t=0$ from an initial speed of $10 \,m/s$ and accelerates uniformly with $2 \,m/s^2$ on a straight road for time $0 \leq t \leq 10 \,s$. Let $S_1$ and $S_2$ be the distance covered by the car in time $3 \leq t \leq 4 \,s$ and $4 \leq t \leq 5 \,s$ respectively. The ratio $\frac{S_2}{S_1}$ is

$A$ car is moving with velocity $V$. It stops after applying brakes at a distance of $20 \ m$. If the velocity of the car is doubled,then what distance will it cover after applying the brakes (in $m$)?

$A$ driver applies the brakes on seeing the red traffic signal $400 \ m$ ahead. At the time of applying brakes,the vehicle was moving with $15 \ m/s$ and retarding at $0.3 \ m/s^2$. The distance of the vehicle from the traffic light one minute after the application of brakes is: (in $m$)

$A$ particle starts from the origin at time $t=0$ and moves in the positive $x$-direction. Its velocity $v$ varies with time as $v=10t \text{ cm/s}$. The distance covered by the particle in $8 \text{ s}$ will be: (in $\text{ cm}$)

An electron starting from rest has a velocity that increases linearly with time,given by $v = kt$,where $k = 2 \, m/s^2$. The distance travelled in the first $3 \, s$ will be ........... $m$.