Distance, Acceleration, Time, and Power

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

So...I did... P=mad/t=ma*(V1t+.5at^2) = 23985

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... :eek:
 

lanedance

Homework Helper
3,305
2
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
 
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
3,305
2
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
 
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?
 
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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