a car is moving with a speed of 36 km/h is retarded by 2m/s. find the time taken to stop the car
Answers
AnswEr :
Given that car moving with a speed of 36 km/h is retarded by 2 m/s².
We have to find time taken to stop the car.
From the data :
Initial speed (u) = 36 km/h
= 36 × (5/18) m/s
= 10 m/s
Retardation (r) = 2 m/s²
Acceleration (a) = -2 m/s²
Final speed (v) = 0 m/s
Now using 1st equation of motion i.e. v = u + at,
Substituting values,
→ 0 = 10 + (-2)t
→ 0 = 10 - 2t
→ 2t = 10
→ t = 10/2
→ t = 5 s
Time taken to stop the car = 5 seconds. (Answer) ✔
Answer :
Given -
- Speed of the car = 36 km/hr
- Retardation = 2 m/s
To Find -
- Time taken to stop the car = ?
Solution -
Initial Speed given is 36 km/hr.
⇒ 36 * 5/18
⇒ 2 * 5
⇒ 10 m/s
Final Speed = 0 m/s
Using Newton's first equation of motion :
⇒ v = u + at
⇒ 0 = 10 + (-2)*(t)
[Retardation is 2 m/s, so acceleration = -2 m/s]
⇒ 0 = 10 - 2t
⇒ -10 = -2t
⇒ t = -10/-2
⇒ t = 5 seconds
Hence, time taken to stop the car is 5 seconds.