Question

A race car accelerates from 21.1 m/s to 57.2 m/s in 2.345 seconds. Caculate its acceleration. (Round to the nearest hundredth)

Answer 1

acceleration is defined as rate of change in velocity

so it is given as

$a = \frac{v_f - v_i}{t}$

now here we know that

$v_f = 57.2 m/s$

$v_i = 21.1 m/s$

$t = 2.345 s$

now we will use the above equation

$a = \frac{57.2 - 21.2}{2.345}$

$a = 15.4 m/s^2$

so here it will accelerate at rate of 15.4 m/s^2

Answer 2

v₀ = initial velocity of the race car = 21.1 meter/second

v = final velocity of the race car = 57.2 meter/second

t = time for which the race car accelerates = 2.345 second

a = acceleration of the race car

acceleration of the race car is given using kinematics equation as

a = (v - v₀)/t

inserting the values

a = (57.2 - 21.2)/2.345

a = 36/2.345

a = 15.4 meter/second²