Q123 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Physics

Consider that a truck is moving initially with $54 \mathrm{~km} / \mathrm{h}$. It has stopped by the driver after looking at an obstacle with a deceleration of $10 \mathrm{~m} / \mathrm{s}^2$. The distance travelled by truck before coming to rest is

A
$12 \mathrm{~m}$
Correct answer
B
$11.25 \mathrm{~m}$
C
$11.30 \mathrm{~m}$
D
$11.20 \mathrm{~m}$

Solution

Given, initial velocity of truck, $u=54 \mathrm{~km} / \mathrm{h}$ $=54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s}$ Final velocity, $v=0$ Deceleration, $a=-10 \mathrm{~m} / \mathrm{s}^2$ As we know that, $v^2=u^2+2 a s$ $\Rightarrow \quad(0)^2=(15)^2+2 \times(-10) s \Rightarrow 20 s=(15)^2$ $s=\frac{15 \times 15}{20}=11.25 \mathrm{~m}$

Hence, the correct answer is option: 'B'.

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