Question

Brakes are applied to a car moving with 20m/s so that it stops in 4 seconds. Find the distance covered

Answer 1

Answer:

40m

Explanation:

v^2 - u^2 =  2as ---->2

v = u + at  --->1

u = 20 m/s

v = 0 (as the car stops)

t = 4s

substitute in eq 1

0 = 20 + 4a  => a = -5 m/s^2

{note that 'a ' is negative here and that is why the car is slowing down (retardation)}

now substitute in eq 2

0^2 - (20)^2 = 2*(-5)*s

-400 = -10s

Therefore s = 40m

Answer 2

Answer

Distance covered = 40 m  $\pink{\bigstar}$

Given

Brakes are applied to a car moving with 20 m/s so that it stops in 4 seconds

To Find

Distance covered

Formula Applied

$\pink{\bigstar}$  v² - u² = 2as

where ,

v denotes final velocity

u denotes initial velocity

a denotes acceleration

s denotes distance

Solution

Initial velocity , u = 20 m/s

Final velocity , v = 0 m/s

Since brakes applied

Time , t = 4 s

Apply 1st equation of motion ,

⇒ v = u + at

⇒ 0 = 20 + a(4)

⇒ 4a = -20

⇒ a = -5 m/s²  $\pink{\bigstar}$

Note : Negative sign denotes retardation

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (0)² - (20)² = 2(-5)s

⇒ -400 = -10s

⇒ 10s = 400

⇒ s = 40 m  $\pink{\bigstar}$

So , Distance covered = 40 m