Question

A car starts moving with a uniform acceleration from its position of rest and it moves $25m$ in $3s$. On applying brakes, it stops after covering a distance. What is the value of d, if its deceleration is $\frac{50}{9}$m s^-2?

• A. $\frac{50}{9}$
• B. $25$
• C. $\frac{50}{3}$
• D. $50$

Answer 1

The correct option is B

$25$

Step 1: Given Data

Initial velocity= $0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Covers $25m$ in $3seconds.$

Let initial acceleration be m s^-2

Step 2: Calculation

We know, For calculating distance,

s =$ut+\frac{1}{2}a{t}^2$

Now putting the above data,

25 = $0x3+\frac{1}{2}\times a\times {\left(3\right)}^2$

a =$\frac{50}{9}m{s}^{-2}$

So final velocity at time 3 seconds will be,

v = u + at

$v=0+$$\frac{50}{9}\times 3$

v = $\frac{50}{3}m{s}^{-1}$.

Now, deceleration = $\frac{50}{9}$m s^-2

Step 3: Final solution

So displacement before it stops given by,

$v^2=u^2-2as$

Since final velocity $v=0m{s}^{-1}$

and initial velocity = $\frac{50}{3}m{s}^{-1}$

$$\begin{gathered} 0^2={\left(\frac{50}{3}\right)}^2-2\times \frac{50}{9}\times s \\ \frac{50\times 50}{9}=2\times \frac{50}{9}\times s \\ s=25m. \end{gathered}$$

Hence the correct option is (B) $25m.$