Posted by **kia** on Thursday, June 30, 2011 at 6:09pm.

Suppose that a car starts from rest at position -3.31m and accelerates with a constant acceleration of 4.15m/s^2. At what time t is the velocity of the car 19.2m/s?

- Physics-Math -
**lola**, Thursday, June 30, 2011 at 6:14pm
v=u+at

19.2=0+4.15t

t=19/4.15

t=4.8

check the answer

- Physics-Math -
**kia**, Thursday, June 30, 2011 at 6:17pm
What is the "u" variable standing for?

- Physics-Math -
**kia**, Thursday, June 30, 2011 at 6:17pm
Your answer's wrong, btw, sorry :(

- Physics-Math -
**lola**, Thursday, June 30, 2011 at 6:21pm
u is the initial velocity that is 0 as it as rest ..

whats the answer ?

- Physics-Math -
**Zero**, Tuesday, August 25, 2015 at 10:27pm
Thread necro... but I wanted to make sure that anyone seeing this can correctly divide 19.2/4.15 = 4.626506024

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