The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hour (mph). At full power, the car can accelerate from zero to 29.0mph in time 1.50s .

Part A

At full power, how long would it take for the car to accelerate from 0 to 58.0mph ? Neglect friction and air resistance. THE ANSWER IS "6 seconds" according the mastering physics.. BUT WHY Please show how you derive that answer...

Part B

A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0mph in time 1.50s , how long would it take to go from zero to 58.0mph ?

A) 6 secs

B) 3 secs

Explanation:

A) CONSTANT POWER

At full power, we have:

Initial velocity = 0mph

Final velocity = 29mph

t = 1.5secs

We are told that the power is constant. Hence, we first find the power applied by the engine of the car:

P = E/t

Where P = power

E = Energy

t = time (in seconds)

Kinetic Energy (E) is given as:

E = 0.5*m*v²

We are not given the mass, so:

P = (0.5*m*v²)/t

P = (0.5 * m * 29²)/1.5

P = 280m J/s

This power is constant but now the final velocity is 59 mph. We can find time by making t subject of the power formula:

P = (0.5*m*v²)/t

t = (0.5*m*v²)/P

t = (0.5*m*58²)/(280m)

t = 6 secs.

B) CONSTANT FORCE

In this case, force is constant:

F = ma = constant

Where a = acceleration.

We can find acceleration, a, by using one of the equations of motion:

a = (v - u)/t

In hours,

t = 1.5/3600 = 0.0004167 hr

a = (29 - 0)/(0.0004167)

a = 69600 mph²

Since the force is constant and mass is constant, that means the acceleration is also constant. Hence, when the final velocity is 58mph, t will be:

a = (v - u)/t

=> t = (v - u)/a

u = 0mph

=> t = 58/69600

t = 0.000833 hr = 3secs