Question

A car changes its velocity from 19.4 m/s northward to 37.5 m/s northward over a time period of 1.33 min. What is the acceleration of the car.

Answer 1

Acc= 0,22m/s^2

Step-by-step explanation:

Remember that acceleration is the change in velocity over a certain period of time, in most of the cases is measured in m/s*2, to calculate acceleration you need to use the next formula:

`Acc=(Vf-Vo)/(t) \\`

Now we convert 1,33 mins to seconds:

1,33*60=79,8s

So if you have this formula we just insert the values into the formula:

`Acc=(37,5m/s-19,4m/s)/(79,8s) \\Acc=(18,1)/(79,8s)\\Acc=0,22m/s^2`

So the acceleration of the car would be 0,22m/s^2

Answer 2

`a=0.22m/s^2`

Step-by-step explanation:

To find the acceleration we need to solve the following formula:

`a=(v_(2)-v_(1))/(t_(2)-t_(1))`

Since the time interval is 1.33 min

`t_(2)-t_(1)=1.33min*(60s)/(1min)=80s`

`a=((37.5-19.4)m/s)/(80s)=0.22m/s^2`