Question

A truck driver driving at 72 km/h finds a red light 100m ahead of him. He instantly applies brakes to stop the truck. The truck retards uniformly ad stops just at the stop line of red light. How much time did the driver take to stop the truck?

Answer 1

Answer: 10 s

Given,

Initial velocity = u = 72 km/m = 20 m/s.

Final velocity = v = 0.

Let the retardation be a m/s sq.

We have s = distance = 100 m  

From the equation of motion,

$v^{2} =u^{2} - 2as$

[on putting the value]

$0 = 20^{2} - 2a*100$

From above we get, a = 2 m/s sq.

Let the truck took t seconds to stop at stop line.

Then,

V = u – at

0 = 20 – 2t

T = 10 s.

Answer 2

Answer:

10 seconds.

Explanation:

here, given that the truck is driving at a speed of 72 km/hr.

⇒ u = 72 km/h

Convert it to SI unit, m/s

⇒ u = 72 km/hr = 72 × 5/18 m/s = 20 m/s

now, it stops just at the stop line, so s = 100 m

now, since it finally comes to rest, v = 0 m/s

To find time, we would need to find acceleration also.

Apply the third equation of motion to find acceleration

⇒ v² - u² = 2as

⇒ 0 - (20)² = 2a × 100

⇒ - 400 = 200a

⇒ a = - 400/200

⇒ a = -2 m/s²

Now, put the value of a, v and u in first equation of motion to get time

v = u + at

0 = 20 + (-2)t

⇒ 0 = 20 - 2t

⇒ - 20 = - 2t

⇒ 20 = 2t (cancelling negative sign)

⇒ t = 20/2

⇒ t = 10 seconds.

Hence, the time that the driver had taken to stop the truck was 10 seconds.