A car accelerates uniformly from 18 km h to 36 km h in 5 min calculate acceleration and distance cover by car in that time
Answers
Given :
▪ Initial velocity = 18kmph
▪ Final velocity = 36kmph
▪ Time interval = 5min
To Find :
▪ Acceleration of car.
▪ Distance covered by car in the given interval of time.
Concept :
☞ Acceleration is defined as the rate of change in velocity.
☞ It is a vector quantity.
☞ It has both magnitude as well as direction.
☞ It can be positive, negative or zero.
☞ SI unit : m/s²
☞ Since, acceleration has said to be constant throughout the motion we can apply third equation of kinematics to calculate distance.
Conversion :
↗ 1kmph = 5/18mps
↗ 18kmph = 18×5/18 = 5mps
↗ 36kmph = 36×5/18 = 10mps
↗ 5min = 5×60 = 300s
Calculation :
❇ Acceleration of car :
→ a = (v - u)/t
→ a = (10 - 5)/300
→ a = 5/300
→ a = 0.017 m/s²
❇ Diatance covered by car :
→ v² - u² = 2as
→ 10² - 5² = 2(0.017)s
→ 100 - 25 = 0.034s
→ s = 75/0.034
→ s = 2205.88 m ≈ 2.2km