A car with speed 72km/hr suddenly applies break. The break has maximum ability to decelerate with
5ms-2. Find time taken to stop the car after applying breaks?
Answers
Answer:
The time taken (t) to stop the car is 4 seconds
Given:
1. Initial velocity (u) = 72 Km/hr.
2. Deceleration (a) = - 5 m/s²
3. Final velocity (v) = 0 m/s
Explanation:
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First we need to convert Velocity into m/s (S.I Units)
we know,
⇒ 1 Km/h = 5 / 18 m/s
⇒ 72 Km/h = 72 × 5 / 18
⇒ 72 Km/h = 4 × 5
⇒ 72 Km/h = 20
⇒ v = 20 m/s
∴ We got the velocity in S.I units.
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From First kinematic equation,
⇒ v = u + a t
Substituting the values,
⇒ 0 = 20 + (-5) × t
⇒ 0 = 20 - 5 × t
⇒ 0 = 20 - 5 t
⇒ 5 t = 20
⇒ t = 20 / 5
⇒ t = 4
⇒ t = 4 sec.
∴ The time taken (t) to stop the car is 4 seconds.
Note:
Symbols have their usual meanings
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Answer: 14.4 seconds
Explanation:
Initial velocity(u) = 72km/h
Final velocity (v) = 0km/h (since brakes are applied)
Acceleration (a) = 5m/s
WE KNOW,
a=v-u/t
⇒ t=v-u/a
⇒t= 0-72/-5
⇒t= -72/-5
⇒t= 14.4 s (ANS)