Asked by kia

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?
Answered by lola
v=u+at
19.2=0+4.15t
t=19/4.15
t=4.8
check the answer
Answered by kia
What is the "u" variable standing for?
Answered by kia
Your answer's wrong, btw, sorry :(
Answered by lola
u is the initial velocity that is 0 as it as rest ..
whats the answer ?
Answered by Zero
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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