Question

A toy car of mass $500g$ travels with a uniform velocity of $25cm{s}^{-1}$ for $5s$. The brakes are then applied and the car is uniformly retarded and comes to rest in further $10s$. Calculate the distance which the car travels after the brakes are applied.

• A. $1.25cm$
• B. $125m$
• C. $1.25m$
• D. $25m$

Answer 1

The correct option is C $1.25m$

Given that this toy car has a velocity of $25cm/sec$ which is also equal to $0.25m/sec$. Here toy car is retarded confirming it has final velocity $0m/sec$ and it came to rest within $10seconds$. We know that acceleration is nothing but rate of change of velocity.
$a=\frac{v-u}{t}$
$a=\frac{0-0.25}{10}$
$a=0.025m/s^2$.
$\frac{v^2-u^2}{a}=s$
${0.25}^2-0^2=2\times 0.025\times s$
$s=1.25m$
Hence displacement is $1.25m$.