Question

A car moving with uniform acceleration covers 450 m in a 5 second interval, and covers 700 m in the next 5 second interval. The acceleration of the car is

• A. $7\frac{m}{s^2}$
• B. $50\frac{m}{s^2}$
• C. $25\frac{m}{s^2}$
• D. $10\frac{m}{s^2}$

Answer 1

The correct option is D

$10\frac{m}{s^2}$

Method I:

$t=5s$
$s_5=450m$
$t=10s$
$s_{10}=450+700=1150$
$s=ut+\frac{1}{2}a{t}^2$
$450=5u+\frac{1}{2}a\left(25\right)...\left(i\right)$
$1150=10u=\frac{1}{2}a\times 100...\left(ii\right)$
Solving equations (i) and (ii), we get
$u=65\frac{m}{s}$
$a=10{\frac{m}{s}}^2$

Let u be the velocity at the start of the first 5 second interval and v be the velocity at the start of next 5 second interval.

From I-equation of motion;
$v=u+at$
$v-u=at...\left(i\right)$
from II-equation of motion;
$S=ut+\frac{1}{2}a{t}^2$

For the first 5 sec. interval,
$S_1=ut+\frac{1}{2}a{t}^2...\left(ii\right)$
For next 5 sec interval,
$S_2=vt+\frac{1}{2}a{t}^2...\left(iii\right)$
$\therefore S_2-S_1=(v-u)t$
Also;
$S_2-S_1=at\times t$ [from equation (i)]
$S_2-S_1=a{t}^2$
$700-450=a\times 5\times 5$
$250=a\times 5\times 5$
$a=10{\frac{m}{s}}^2$