# Problem on distance, time and velocity when acceleration is the only given.

1. Dec 11, 2008

### technowar

Two cars, car A and car B. Car A and car B initially at rest, both starts at the same time. Car A is initially at some distance away behind car B.

A.) How long does it take for car A to overtake car B?
B.) How far was car A behind car B initially?
C.) What is the speed of each car just before overtaking?

Given:
acar A= 3.5 m/s2
acar B= 2.2 m/s2
tcar A=tcar B because they started on the same time.
vi of car A=vi of car B=0 because they are initially at rest.

Please someone help me with this?

Last edited: Dec 11, 2008
2. Dec 11, 2008

### technowar

Please check my solutions regarding another problem.

Boy A is on the roof, 46m above the ground. Boy B is sneaking slowly alongside the tower at a constant speed of 1.2m/s. How far should boy B from the tower when boy A releases the stone from the roof to make a sure hit?

Given:
Boy A on tower
vi=0
yi=0
yf=46m
g=-9.8m/s2

Boy B alongside
xi=0
v=1.2m/s
a=0 because boy B has a constant speed.

Solutions:
To solve for time t, I used the formula yf=vit-1/2gt2. y=-1/2gt2
-2y=gt2
-2y/g=t2
sqrt(t2)=sqrt(-2y/g)
t=sqrt[-2(46m)/(-9.8m/s2)]
t=3.06399437s

Next question, How far should boy B from the tower when boy A releases the stone from the roof to make a sure hit? I used the formula xf=vit+1/2at2.
xf=(1.2m/s)(3.06399437s)
xf=3.676733244m

Do I get it right? Please make some corrections if I do it the wrong way. Thank you again.