Question

What will be the solution​

Answer 1

Let the reaction time of the man be $t_{r}$.

Here's what's happening:

First, some event occurs - the man takes $t_{r}$ time to notice that event - in this time, the car is travelling at a constant speed. Then, the man applies brakes to slow down his vehicle, which causes the vehicle to decelerate at a constant rate, and come to a stop.

Let's convert the given speeds to m/s, for convenience:

72 kmph = 20 m/s ($u_{1}$)

36 kmph = 10 m/s ($u_{2}$)

54 kmph = 15 m/s ($u_{3}$)

First, let's remove the extra distance traveled by the car, due to the man's reaction time:

$S_{1} = 30 - 20t_{r}$

$S_{2} = 10 - 20t_{r}$

Now, applying equation of motion $v^2 = u^2 + 2aS$

Let acceleration be -a (that is, deceleration = a).

v in each case is 0 m/s (stoppage).

So: $0 = u^2 - 2aS$

$u^2 = 2aS$

Putting $S_{1}, u_{1}$ and $S_{2}, u_{2}$:

1. $400 = 2 * a * (30 - 20t_{r})$

$20 = a * (3 - 2t_{r})$

2. $100 = 2 * a * (10 - 20t_{r})$

$5 = a * (1 - 2t_{r})$

Trying to equate the two equations above, we get:

$a * (3 - 2t_{r}) = 4 * a * (1 - 2t_{r})$

Cancelling a on both the sides:

$3 - 2t_{r} = 4 - 8t_{r}$

$6t_{r} = 1$, or, $t_{r} = \frac{1}{6}s$

And putting this value in either equation, we get:

$a = \frac{15}{2}$

Finally, let's find out the thing being asked:

Firstly, the man will take $t_{r}$ time to react. The distance covered during this time:

$d_{1} = 15 * (\frac{1}{6}) = \frac{5}{2}m$ (since u = 15 m/s)

Now, he'd apply the brakes, and the car will decelerate at a rate of 7.5 $m/s^2$. Apply the equation of motion to get this distance:

$15^2 = 2 * \frac{15}{2} * d_{2}$

$d_{2} = 15m$

The net distance covered this time = $d_{1} + d_{2}$ = 2.5m + 15m = 17.5m.