# Answer to Question #717 in Mechanics | Relativity for Rooky

Question #717
If a car starts from rest and accelerates at 5.0 m/sec^2 for 6.0 seconds, how far did it travel and how fast was it moving at the end of the 6.0 seconds?

1
2010-10-03T04:36:54-0400
The motion of the car in that 6 seconds have been a motion with constant acceleration. One of the
equations of motion of a body in constant acceleration is:
a=&Delta;v/&Delta;t&rArr;&Delta;v=a*&Delta;t
&rArr;v-v0=a*&Delta;t&rArr;v=a*&Delta;t+v0
Where v is the final velocity and v0 is the initial velocity and v0 = 0. Now with all the data we have, we can get for the final velocity:
v=a*&Delta;t=5*6=30m/s
For the distance the car traveled in that time, we use another one of the equations of motion of a body in constant acceleration, namely the time-independent one:
v&sup2;-v&sup2;0=2a*&Delta;x&rArr;&Delta;x=(v&sup2;-v&sup2;0)/2a
and since v0 = 0 we get &Delta;x=v&sup2;/2a
and using the data we have, we get &Delta;x=v&sup2;/2a=30&sup2;/2*5=900/10=90m
We could nd the distance by another method, using another equation of motion:
Delta;x=x-x0=1/2a*t&sup2;+v0*t=1/2*5*6&sup2;+0=90m

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