Question

An athelete takes 2 secs to reach his maximum speed of 18 km/hr . what is the magnitude of his average acceleration? Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution :

Initial velocity=u= 0 m/s
Final velocity =v=18km/hr=18x5/18=5 m/s
time= 2sec
Average acceleration =a=?

Formula to be used : a= v-u/t

a=5-0/2 =2.5 m/s²
∴ Magnitude of his average acceleration is 2.5m/s²

Answer 2

Answer:

The magnitude of his average acceleration is 2.5 m/s²

Explanation:

Given that,

Initial velocity u= 0 m/s

Final velocity v = 18 km/hr=5 m/s

Time t = 2 sec

The formula of the average acceleration is

$a =\dfrac{v-u}{t}$

Where, v = final velocity

u = initial velocity

t = time

Put the value of u,v and t in the formula

$a = \dfrac{5-0}{2}$

$a = 2.5 m/s^2$

Hence, The magnitude of his average acceleration is 2.5 m/s²