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Chapter 1: Q51E (page 62)

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ay = (2.80 m/s3) t, where the +y-direction is upward. (a) What is the height of the rocket above the surface of the earth at t = 10.0 s? (b) What is the speed of the rocket when it is 325 m above the surface of the earth?

Short Answer

Expert verified

(a) Height of the rocket above the surface of the is 466.66 m.

(b) Speed of the rocket is 109.89 m/s2

Step by step solution

01

Identification of given values

a(t)=2.80t

Integrating above equation

vt=1.4t2+c

Here c is the constant.

At t=0 , v=0 so that c=0

vt=1.4t2

Again integrating

xt=1.4t33+d

Again d is the constant.

At t=0 , x(height)=0 so that d=0

xt=1.4t33

Now, t = 10 s

xt=1.4脳1033xt=466.66m

02

calculation for the speed

325=1.4脳t33t=8.86sNowvt=1.4t2vt=1.4脳8.162vt=109.89m/s2

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