Question

a car is rolling backward when it hits the gas. after 8.25 s it is moving forward at 8.62m/s, and is 12.9m to the right of its starting pount. what was its starting velocity ?

Answer 1

-5.49 m/s

Step-by-step explanation:

I just had this same question on my Acellus Academy physics course for 9th grade. Hope this helps ;)

Answer 2

Step-by-step explanation:

The given data is as follows.

Final velocity (v) = 8.62 m/s, Initial velocity (
`v_(o)`) = ?

time = 8.25 sec, distance (s) = 12.9 m

Hence, formula to calculate the initial velocity is as follows.

v =
`v_(o) + a * t`

8.62 m/s =
`v_(o) + a * 8.25 s`

`v_(o)` = 8.62 m/s - 8.25a ......... (1)

Also, s =
`v_(o) * t + (1)/(2)a * t^(2)` ...... (2)

Therefore, substitute the value of
`v_(o)` from equation (1) into equation (2) as follows.

s =
`v_(o) * t + (1)/(2)a * t^(2)`

s =
`8.62 m/s - 8.25a * t + (1)/(2)a * t^(2)` ....... (3)

Now, putting the values of s and t into equation (3) as follows.

s =
`8.62 m/s - 8.25a * t + (1)/(2)a * t^(2)`

12.9 m =
`(8.62 m/s - a * 8.25sec) * 8.25 s + (1)/(2)a * (8.25 sec)^(2)`

a = 1.71
`m/s^(2)`

Therefore, using equation (1) the value of initial velocity will be as follows.

`v_(o)` =
`8.62 m/s - 8.25 sec * a`

=
`8.62 m/s - 8.25 sec * 1.71 m/s^(2)`

= -5.49 m/s

Thus, we can conclude that starting velocity of the car is -5.49 m/s.