# Distance, Acceleration, Time, and Power

#### Norngpinky

1. The problem statement, all variables and given/known data
If a 1500kg car can accelerate from 35 km/h to 55 km/h in 4.0 s, how long will it take to accelerate from 55 km/h to 75 km/h? Assume the power stays the same, and neglect frictional losses.

2. Relevant equations
Converting the velocities to m/s, I got... V1=9.7m/s, V2=15m/s, V3=21m/s

Power = work/time = force*distance/time = mass*acceleration*distance/time

Acceleration1 = (V2-V1)/T1 (which is 4 sec) = 1.3 m/s^2

I'm using d = vt + (1/2)at^2 to find distance <----- This I am not totally sure off...

3. The attempt at a solution

P1 = P2 as the problem stated...

So... a2 = (v3-v2)/t...After substitution I have

a = P / (m*(v3-v2)) = 1.13 m/s^2

Using a = (v3-v2)/t...I solved for t... I got 6.8s

It says I got the wrong answers, but can't figure out there I did it wrong...

#### lanedance

Homework Helper
Hi Norngpinky

are you using integrals in this yet?

As noted P = mav is constant, (note v=d/t) so you could figure out how to include this in your differential equations & integrate

however you could also note that Power = Energy/time

do you know the equation for kinetic energy? If we assume all the power goes into creating kinetic energy & use the fact the power stays constant, this means the change in kinetic energy divided by time will be constant

#### Norngpinky

Hi Norngpinky

are you using integrals in this yet?

As noted P = mav is constant, (note v=d/t) so you could figure out how to include this in your differential equations & integrate

however you could also note that Power = Energy/time

do you know the equation for kinetic energy? If we assume all the power goes into creating kinetic energy & use the fact the power stays constant, this means the change in kinetic energy divided by time will be constant
Oh yes... I don't know why I didn't think of that. I know I messed up on the P=mav since P can be Fv (average) as well. . . and didn't know how to get average v...Did the addition of velocities divided by two, but still got the wrong answer.

Thank you ^^ Your explanation makes a lot of sense.

#### lanedance

Homework Helper
yeah you will have a problem with as the way you are trying to do it really requires integral calculus

try the energy way

P = (E1-E2)/dt - we know this is constant

andthe energy is all going into kinetic energy
E1 = (1/2)*m*(v1)^2

So use this for your first case to find what P is

then as you know the starting velocity and time of acceleration you should be able to solve for final velocity using the P calculated previously

#### baha jaff

An airplan moves on the runway of the airport at aconstant acceleration,then it takes off after it travels 1.5 km on the runway at avelocity of 210 km/h . what is the acceleration of airplane?

#### INeedHelpPls

using v squared = u squared + 2as, a = (v squared - u squared)/2s where v= final velocity u= initial velocity s= displacement a = accel

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