# A Car with Constant Power

1. Mar 17, 2009

### raging11

1. The problem statement, all variables and given/known data
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 30.0 mph in time 1.30s .

At full power, how long would it take for the car to accelerate from 0 to 60.0 mph ? Neglect friction and air resistance.

2. Relevant equations

No clue

3. The attempt at a solution
I assumed that since it takes 1.3s to go from 0-30, it would take 2.6 to got from 0-60, yet it was wrong and got this for a feedback. :
Think about how kinetic energy is related to velocity, and how power is related to energy.

2. Mar 17, 2009

### jdougherty

The feedback you got should give you a clue as to what the relevant equations are. Do you know of some equations which relate kinetic energy to velocity and power to energy?

3. Mar 17, 2009

### raging11

KE=1/2mv^2 and Power=??

4. Mar 17, 2009

### LowlyPion

Welcome to PF.

Constant power is constant watts.

What is a watt? J/s

What is a joule?

well ... it's the units of ½mv² for one thing.

5. Mar 18, 2009

### bjd40@hotmail.com

power = force(instantenuous) X speed (instantenuous)

6. Mar 18, 2009

### raging11

Im still stumped! I have no clue how to transfer from knowing 1.3s at Full power to go from 0-60

7. Mar 18, 2009

### leah_91

I would figure out the acceleration of the car first using the 0-30 in 1.3 sec. then use the known acceleration (since it would be constant at full power) to find out how long it would take the car to go 0-60 with that acceleration.

8. Mar 19, 2009

### LowlyPion

Think of power then as something that is added at a constant rate. It's energy after all and as it collects it becomes kinetic energy of your car.

If it took one bucket of energy to get you to 1/2mv2 then how many buckets of energy will it take to get you to 1/2m(2v)2 ?

Since the buckets are added in this case at 1.3s per bucket - a constant rate -...