Question

The driver of a car moving at a speed of $10\text{m/s}$ applies the brakes after noticing an obstacle. The retardation due to the brakes is $4{\text{m/s}}^2$. What is the distance travelled by the car till it stops?

• A. $12\text{m}$
• B. $12.5\text{m}$
• C. $10\text{m}$
• D. $11\text{m}$

Answer 1

The correct option is B $12.5\text{m}$

We know,
$u=10\text{m/s}$
$a=-4{\text{m/s}}^2$

Let's say the car stops at $t\text{s}$
At $t\text{s}$, $v=0$ (because the car stopped)
Using the 1st equation
$v=u+at$
We get,
$\Rightarrow 0=10+(-4)t$
$\Rightarrow 4t=10$
$\Rightarrow t=\frac{10}{4}=2.5\text{s}$

To get the distance travelled by the car we can use the 2nd equation of motion
$s=ut+\frac{1}{2}a{t}^2$
Replacing values of $u$, $a$ and $t$ in this equation, we get
$\Rightarrow s=ut+\frac{1}{2}a{t}^2=(10\times 2.5)-\frac{1}{2}\times 4\times {2.5}^2$
$\Rightarrow s=12.5\text{m}$