Question

a car is going at a speed of 10m/s. suddenly he sees a boy 50m in front. to avoid accident in what time would the car have to stop​

Answer 1

CORRECT QUESTION:

A car is going at a speed of 10m/s. Suddenly he sees a boy 50m in front. In order to avoid accident, in what time would the car have to stop?

Calculation:

First of all, let's find the retardation required:

$\rm {v}^{2} = {u}^{2} + 2as$

$\rm \implies {0}^{2} = {10}^{2} + 2a(50)$

$\rm \implies 2a(50) = - 100$

$\rm \implies 100a = - 100$

$\rm \implies a = - 1 \: m {s}^{ - 2}$

Now, time required be t :

$\rm \: v = u + at$

$\rm \implies \: 0 = 10 + ( - 1)t$

$\rm \implies \: t = 10 \: sec$

So, time required is 10 seconds.

Answer 2

Second Equation Of Motion Formula :

$\begin{gathered}\begin{gathered}\begin{gathered}\mapsto \sf\boxed{\bold{\pink{v²= u² \: + \: 2as}}}\\\end{gathered}\end{gathered}\end{gathered}$

$\rm \implies {0}^{2} = {10}^{2} + 2a(50)$

$\rm \implies 2a(50) = - 100$

$\rm \implies 100a = - 100$

$\rm \implies a = - 1 \: m {s}^{ - 2}$

So, the acceleration us $1 \: m {s}^{ - 2}$

Now, time required be t :

First Equation Of Motion Formula :

$\begin{gathered}\begin{gathered}\begin{gathered}\mapsto \sf\boxed{\bold{\pink{v = u + at}}}\\\end{gathered}\end{gathered}\end{gathered}$

$\rm \implies \: 0 = 10 + ( - 1)t$

$\rm \implies \: t = 10 \: sec$

So, time required is 10 seconds.