Question

a car is moving with a velocity of 72 kilometre per hour on the application of brakes car stops in 10 seconds find 1 retardation 2 distance covered by the car 3 before it stops braking force if mass or car is 800 kg ​

Answer 1

Answer:-

Given:

Initial velocity of the car (u) = 72 km/h = 72 * 5/18 = 20 m/s.

Final Velocity of the car (v) = 0 m/s [ breaks are applied ]

Time (t) = 10 s

Mass of the car (m) = 800 kg

We have to find:

• Retardation
• Distance covered by the car before applying breaks.
• Force.

1) Retardation is nothing but "- acceleration"

We know that,

Acceleration (a) = (v - u) / t

→ a = (0 - 20) / 10

→ a = - 20/10

→ a = - 2 m/s²

Hence the deceleration of the car is - ( - 2) = 2 m/s².

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2) Using second equation of motion,

→ s = u*t + 1/2 * at²

Here acceleration is negative , So the equation becomes:

→ s = u*t - 1/2 a*t² [ s is the Distance covered]

Putting the values we get,

→ s = (20)(10) - 1/2 * (2) * (10)²

→ s = 200 - 100

→ s = 100

Hence, the distance covered by the car before applying breaks is 100 m.

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3) Force = mass * acceleration

→ F = (800) * (- 2)

→ F = - 1600 N

Hence, the force applied by the breaks against the car is 1600 N.

Answer 2

Step-by-step explanation:

1) Retardation:

Using first equation of motion,

v = u + at

Given; u is 72 km/hr or 20 m/s, v is 0 m/s and t is 10 sec.

→ 0 = 20 + a(10)

→ a = -2

Hence, the Retardation is 2 m/s².

2) Distance:

Using the second equation of motion,

s = ut + 1/2 at²

→ s = 20(10) + 1/2 (-2)(10)²

→ s = 200 - 1(100)

→ s = 100

Hence, the distance covered by the car is 100 m.

3) Force:

Force = Mass × Acceleration

Given: Mass is 800 kg and acceleration is -2 m/s².

→ F = 800(-2)

→ F = -1600N

Hence, the force is 1600N.