Question

a car traveling at speed 72km/hr suddenly applies the brake with the deceleration of 5m/s^2. find the stopping distance of thr car.​

Answer 1

Answer:

40 m

Explanation:

Given - a car travelling at a speed of 72km/hr, i.e. 20m/s

         - deceleration is 5m/s^2

Now, from third equation of motion,

v^2-u^2 = 2as (s = distance, a = negative acceleration),

v = 0 m/s(car comes to rest)

u = 20m/s

0 - 20^2 = 2x(-5)xs

=>-400 = -10s

=>s = 40m

Therefore, the stopping distance of the car is 40m

I hope this helps, good night

Answer 2

Given that, the initial velocity (u) of the car is 72 km/hr. (72 × 5/18 = 20 m/s)

Also given that, when sudden brakes are applied then the deceleration of 5m/s². (a = -5 m/s²)

Final velocity (v) = 0 m/s

We have to find the stopping distance (s) of the car.

Using the third equation of motion.

v² - u² = 2as

→ (0)² - (20)² = 2 × (-5) × s

→ - 400 = -10s

Divide with 10 on both sides

→ -400/10 = -10/10 × s

→ -40 = -s

→ s = 40

Therefore, the stopping distance of the car is 40 m.