A car starts from rest, attains a velocity of 36 KMPH with an acceleration of 0.2 m/s2 travels
9 km with this uniform velocity and then comes to rest with a uniform deceleration of 0.1 m/s2. Find the total time of travel by the car.
Answers
total time travel by the car is 17.5 min
It has given that a car start from rest i.e., u = 0 , attains a velocity v = 36 km/h = 10 m/s with an acceleration, a = 0.2 m/s² travels 9 km with uniform velocity and then comes to rest with uniform deceleration, b = -0.1 m/s².
we have to find the total time of travel by the car.
case 1 : using formula, v = u + at
⇒10 = 0 + 0.2 × t₁
⇒t₁ = 50 sec
case 2 : s = 9km = 9000 m , v = 10 m/s
so, t₂ = s/v = 9000/10 = 900 sec
case 3 : here, u' = v = 10 m/s, v' = 0, deceleration, b = -0.1 m/s²
using formula, v = u + at
⇒0 = 10 + (-0.1)t₃
⇒t₃ = 100 sec
total time travel by the car = t₁ + t₂ + t₃
= 50 + 900 + 100 = 1050 sec = 17.5 min
therefore, total time travel by the car is 17.5 min
Answer:
Explanation:
case 1 : using formula, v = u + at
⇒10 = 0 + 0.2 × t₁
⇒t₁ = 50 sec
case 2 : s = 9km = 9000 m , v = 10 m/s
so, t₂ = s/v = 9000/10 = 900 sec
case 3 : here, u' = v = 10 m/s, v' = 0, deceleration, b = -0.1 m/s²
using formula, v = u + at
⇒0 = 10 + (-0.1)t₃
⇒t₃ = 100 sec
total time travel by the car = t₁ + t₂ + t₃
= 50 + 900 + 100 = 1050 sec = 17.5 min