Question

A rocket traveling at 36.96 m/s is accelerated uniformly to 74.44 m/s over a 21.37 s interval. What is its displacement during this time?

Answer 1

s = 1190.3 m = 1.19 km

Step-by-step explanation:

First we find the acceleration by using first equation of motion:

Vf = Vi + at

where,

Vf = Final Velocity = 74.44 m/s

Vi = Initial Velocity = 36.96 m/s

a = acceleration = ?

t = time = 21.37 s

Therefore,

74.44 m/s = 36.96 m/s + a(21.37 s)

a = (74.44 m/s - 36.96 m/s)/21.37 s

a = 1.75 m/s²

Now, we use 2nd equation of motion to find distance covered (s):

s = Vi t + (1/2)at²

s = (36.96 m/s)(21.37 s) + (1/2)(1.75 m/s²)(21.37 s)²

s = 789.83 m + 400.47 m

s = 1190.3 m = 1.19 km