# Two cars are 175m apart, facing each other. At that instant, the first car is traveling at 9.0 km/h [W] and is accelerating at 3.6m/s^2 [W]. The second car is traveling at a constant velocity of 21.5 m/s [E].?

## a) How long will it take for the cars to hit each other? b) With what speed will the first car be traveling?

(a) $5.236 \setminus s \setminus \quad$ (b) $21.35 \setminus \setminus \textrm{\frac{m}{s}}$

#### Explanation:

Let the cars hit each other after time $t$ from initial state.

The distance ${d}_{1}$ traveled in the time $t$ by the first car moving with initial velocity $u = 9 \setminus \setminus \textrm{k \frac{m}{h} r} = 2.5 \setminus \setminus \textrm{\frac{m}{s}}$ & an acceleration a=3.6\ \text{m/}s^2 is given by second equation of motion,

${d}_{1} = u t + \setminus \frac{a {t}^{2}}{2}$
${d}_{1} = 2.5 t + \setminus \frac{3.6 {t}^{2}}{2}$
${d}_{1} = 2.5 t + 1.8 {t}^{2} \setminus \ldots \ldots \left(1\right)$

The distance ${d}_{2}$ traveled in the time $t$ by the second car moving with constant velocity $v = 21.5 \setminus \setminus \textrm{\frac{m}{s}}$

${d}_{2} = v \setminus \times t = 21.5 t \setminus \ldots \ldots \ldots \left(2\right)$

At the point of collision, the sum of distances ${d}_{1}$ & ${d}_{2}$ traveled by both the cars must be equal to the initial distance $175 \setminus m$

$\setminus \therefore {d}_{1} + {d}_{2} = 175$

$2.5 t + 1.8 {t}^{2} + 21.5 t = 175$

$1.8 {t}^{2} + 24 t - 175 = 0$

$9 {t}^{2} + 120 t - 875 = 0$

Solving the above quadratic equation, as follows
$t = \setminus \frac{- 120 \setminus \pm \setminus \sqrt{{120}^{2} - 4 \left(9\right) \left(- 875\right)}}{2 \left(9\right)}$
$= \setminus \frac{- 120 \setminus \pm 30 \setminus \sqrt{51}}{18}$
But time $t > 0$ hence, we have

$t = \setminus \frac{- 120 + 30 \setminus \sqrt{51}}{18} = \setminus \frac{5 \setminus \sqrt{51} - 20}{3} = 5.236 \setminus s$

Hence the cars will hit each other at time $t = 5.236 \setminus s$

Now, the speed of first car just before collision at time $t = 5.236 \setminus s$ is given by first equation of motion,

$v = u + a t$

$v = 2.5 + 3.6 \setminus \cdot 5.236 = 21.35 \setminus \setminus \textrm{\frac{m}{s}}$