A1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. ignoring friction, how much power did that require?

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .


One interesting thing to notice while you're (I'm) solving this problem
is the fact that although the car's speed only increased by about 47%,
its kinetic energy more than doubled ... increased by about 115%. 
That's because kinetic energy is proportional to the SQUARE of the
speed.

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.

Sorry i don’t know what to do