Question

A braking car on a drying pavement has a constant acceleration of 3.0 m/s^2. If the car was moving at 10.0 m/s, approximately how long would it take to stop?

10 s, 17 s, 30 s, or 3.3 s

@jim_thompson5910 :)

Answer 1

so a=3
t=?
vi=0
and would vf=10 ?
@jim_thompson5910 ?

Answer 2

"If the car was moving at 10.0 m/s, approximately how long would it take to stop?"

the car is initially going 10 m/s (vi)
the stopped speed is 0 m/s (vf)

Answer 3

ohh okay

so vi=10
vf=0
a=3
t=?

so using vf=vi+at again?

Answer 4

0=10+3t ?

Answer 5

yes, solve for t

Answer 6

3t=-10
t=-3.3333....

?

Answer 7

I forgot to mention, but since you're decelerating, the value of 'a' is negative

Answer 8

a = -3

Answer 9

ohhh okay:)

so 0=10+(-3t)
-10=-3t
t=3.3333

so 3.3 s=t ?

Answer 10

yes roughly

Answer 11

and so anytime it's decelerating, it will be -?

and okie yay!

Answer 12

yes 'a' is negative when you decelerate

Answer 13

awesome!!! thanks bunches!! :)