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

The Correct Option is B

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$