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

Answer: = . /

Explanation:

The acceleration is

=   − 0

In our case, the initial velocity has minus sign.

Thus,

= − (−0) = + 0

Substituting

0 = 2 ( + 0 ) − = 2 + 0 2 −

Thus,

0 2 = 2 −

So,

0 = − = 8.62 − 12.9

8.25 = 7.06 m/s

Answer: = . /

Answer 2

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 = $v_{o} + a \times t$

                 8.62 m/s = $v_{o} + a \times 8.25 s$

                $v_{o}$ = 8.62 m/s - 8.25a ......... (1)

Also,   s = $v_{o} \times t + \frac{1}{2}a \times t^{2}$ ...... (2)

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

                s = $v_{o} \times t + \frac{1}{2}a \times t^{2}$

               s = $8.62 m/s - 8.25a \times t + \frac{1}{2}a \times t^{2}$ ....... (3)  

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

         s = $8.62 m/s - 8.25a \times t + \frac{1}{2}a \times t^{2}$  

        12.9 m = $(8.62 m/s - a \times 8.25sec) \times 8.25 s + \frac{1}{2}a \times (8.25 sec)^{2}$  

               a = 1.71 $m/s^{2}$  

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

                      $v_{o}$ = $8.62 m/s - 8.25 sec \times a$

                              = $8.62 m/s - 8.25 sec \times 1.71 m/s^{2}$

                              = -5.49 m/s

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