A car undergoing uniform acceleration of 3 m/s2 changes its velocity from 15 m/s to 30 m/s. Find:
(a) Distance travelled between this time.
(b) Time required to make this change.
Answers
Given :
▪ Initial velocity = 15m/s
▪ Final velocity = 30m/s
▪ Acceleration = 3m/s²
To Find :
➠ Distance covered by car.
➠ Time of journey.
Concept :
⇒ Since, acceleration has said to be constant throughout the motion, we can easily apply equation of kinematics to solve this question.
✪ First equation of kinematics :
➢ v = u + at
✪ Third equation of kinematics :
➢ v² - u² = 2as
where,
◕ v denotes final velocity
◕ u denotes initial velocity
◕ a denotes acceleration
◕ s denotes distance
◕ t denotes time
Calculation :
፨ Distance covered by car :
➳ v² - u² = 2as
➳ (30)² - (15)² = 2(3)s
➳ 900 - 225 = 6s
➳ 675 = 6s
➳ s = 112.5m
፨ Time of journey :
➶ v = u + at
➶ 30 = 15 + 3t
➶ 15 = 3t
➶ t = 5s
Answer:
(a) 112.5 m
(b)5 secs
Explanation:
v^2 - u^2=2as
30^2 - 15^2 = 2×3×s
s = 3×225/6 = 112.5m
v = u + at
(v - u)/a = t
15/3 = 5secs