Question

A car travels 200 km in 2 h and travels 240 km in next 3 h. If the acceleration is constant then the distance it will travel in the next one hour is

• A. 48 km
• B. 64 km
• C. 72 km
• D. 84 km

Answer 1

Answer: (B)

64 km

Answer 2

Let $u$ and $a$ are initial velocity and acceleration of cars, respectively.
$\therefore s =ut +\frac{1}{2}at^{2}$
$s =200, t=2$
$200 = 2u +2a$
$\Rightarrow u +a = 100\quad...\left(i\right)$
$s_{1} +s_{2} = u\left(t_{1}+t_{2}\right) +\frac{1}{2}a\left(t_{1}+t_{2}\right)^{2}$
$s_{1} = 200, s_{2} = 240, t_{1} = 2, t_{2} = 3$
$440 = 5u +\frac{25a}{2}$
$\Rightarrow 10u +25a = 880 ...\left(ii\right)$
From Eqs. $\left(i\right)$ and $\left(ii\right)$, we get $u = 108, a = - 8$
$s_{1} +s_{2} +s_{3} = u\left(t_{1} +t_{2}+t_{3}\right) +\frac{1}{2} a\left(t_{1} +t_{2} +t_{3}\right)^{2}$
$t_{1} = 2, t_{2} = 3, t_{3} = 1, s_{1} = 200, s_{2} = 240, s_{3} = ?$
$440 +s_{3} = 108\left(6\right) - \frac{8}{2}\left(6\right)^{2}$
$s_{3} = 648 -144 -440 = 84 \,km$