Take into account that the average power is given by:
![P=(F\cdot d)/(t)](https://img.qammunity.org/2023/formulas/physics/college/5yipxfbvsaquw8mr4d6ln6o5e53ymv4m66.png)
where F is the force, d is the distance and the t the time.
Then, it is necessary to calculate F. Use the following formula:
![F=m\cdot a](https://img.qammunity.org/2023/formulas/physics/high-school/krcjg3szwdycbclm76qbymeg780wcathgy.png)
m: mass of the car = 2000 kg
a: acceleration = ?
Calculate acceleration as follow:
![a=(v-v_o)/(t)=((0m)/(s)-(30m)/(s))/(25s)=1.2(m)/(s^2)](https://img.qammunity.org/2023/formulas/physics/college/n97ymjbvzm25i8w3uqytldfr71cwlnjg58.png)
where you have considered that the final velocity is 0m/s because the car stops.
Then, the force is:
![F=(2000kg)(1.2(m)/(s^2))=2400N](https://img.qammunity.org/2023/formulas/physics/college/4iwmjv9iy6h58hz83i2q32yfi77f9taf5u.png)
Now, consider that it is necessary to calculate distance d in order to able to calculate P.
Calculate distance d as follow:
![d=(v^2_o)/(2a)=((30(m)/(s))^2)/(2(1.2(m)/(s^2)))=375m](https://img.qammunity.org/2023/formulas/physics/college/lrtsb2uvd61izhx39rfsjgbr5dfa6jwhjs.png)
Then, by replacing d, F and t = 25 s, you obtain for the power:
![P=(2400N\cdot375m)/(25s)=36000W=36kW](https://img.qammunity.org/2023/formulas/physics/college/out38d2feoyzmizb69fd062f86g3cmy5gr.png)
Hence, the average power produced by the brakes was 36kW