a motorbike accelerates uniformly from 54 km/h to 72 km/h in 2 sec. calculate i. acceleration and distance covered by the motorbike in that time
Answers
Concept:
We need to utilize the first and third kinematic equations of motion. They are expressed as- v = u + at and v² = u² + 2as.
Given:
A motorbike accelerates uniformly from initial velocity, u = 54 km/h to final velocity, v = 72 km/h
Time = 2seconds.
Find:
We need to determine the:
i. acceleration, a covered by the motorbike in that time
ii. distance, s covered by the motorbike in that time
Solution:
It has been given that the,
Initial velocity = 54 km/h which is converted to 15 m/s
Final velocity = 72 km/h which is converted to 20 m/s
We know, the first equation of kinematic motion as, v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration while t is the time.
Therefore, a = (v - u)/t
a = 20 - 15/2
a = 2.5 m/s²
For distance, we need to use the third kinematic equation of motion
v² = u² + 2as where s is the distance
s = v² - u²/2a
s = 20² - 15²/ 2(2.5)
s = 35m is the distance covered by the motorbike in that time
Thus, acceleration and distance covered by the motorbike in 2 seconds are 2.5 m/s² and 35m, respectively.
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