Question

a break applied to a car produces an acceleration of 10 metre per second square in the opposite direction to the motion if it takes 5 second to stop after the application of brain then how much distance is covered by the car before coming to rest​

Answer 1

Given:-

• Acceleration ,a = -10m/s² (opposite direction of motion)

• Final velocity ,v = 0m/s
• Time taken ,t = 5s

To Find:-

• Distance covered by the car before coming to rest ,s

Solution:-

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀According to the Question

Firstly we calculate the initial velocity of the car . Using 1st equation of motion

• v = u + at

where,

• v is the final velocity
• a is the acceleration
• u is the initial velocity
• t is the time taken

Substitute the value we get

$:\implies$ 0 = u + (-10) × 5

$:\implies$ -u = -50m/s

$:\implies$ u = 50m/s

So the initial velocity of the car is 50m/s

Now, calculating the final velocity of the car by using 3rd equation of motion.

• v² = u² + 2as

substitute the value we get

$:\implies$ 0² = 50² + 2(-10)× s

$:\implies$ 0 = 2500 -20s

$:\implies$ -2500 = -20s

$:\implies$ 20s = 2500

$:\implies$ s = 2500/20

$:\implies$ s = 125 m

• Hence, the distance covered by the car before coming to rest is 125 metres.

Answer 2

Given :-

A break applied to a car produces an acceleration of 10 metre per second square in the opposite direction to the motion if it takes 5 second to stop after the application of brain

To Find :-

Distance covered

Solution :-

Since the car move in opposite direction so, acceleration will be -10 m/s².

v = u + at

0 = u + -10(5)

0 = u + (-50)

0 - u = - 50

-u = -50

u = 50 m/s

Now

v² - u² = 2as

(0)² - (50)² = 2(-10)(s)

0 - 2500 = -20s

-2500 = -20s

-2500/-20 = s

2500/20 = s

125 = s