Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Car Acceleration question

  1. Oct 31, 2006 #1
    If a 1500 kg car can accelerate from 35 km/h to 55 km/h in 3.8 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

    what I have tried so far:

    a=5.263 km/h squared

    75 squared= 55 squared + 2(5.263squared)d


    -8.067= 2.6315t

    t=3.06 but the answer is obviously wrong
  2. jcsd
  3. Oct 31, 2006 #2


    User Avatar
    Homework Helper

    Since acceleration is constant, you can use [tex]v(t) = v_{0} + at[/tex]. Simply apply that formula to the case when v0 = 55 km/h, and v(t) = 75.
  4. Oct 31, 2006 #3
    diodnt they ask you how LONG it takes it takes to reach 75km/h from 55km/h? That means time

    you have the initial velocity, you have the final velocity, you have your acceleration, cant you find the time WITHOUT having to find distance?
  5. Oct 31, 2006 #4
    The question says assume power is constant; since kinetic energy has [itex]v^2[/itex] in it, it will take longer to up the speed by 20km/h than previously.
  6. Oct 31, 2006 #5


    User Avatar
    Homework Helper

    You're right. I missed the word 'power'. So it becomes a dynamics problem.
    Last edited: Nov 1, 2006
  7. Oct 31, 2006 #6
    when I plug the numbers into the equation 75=55+at

    I get 20=5.263t

    t=3.8001, but when I plug that in it is wrong. And that would be the same time as it took to get from 35 to 55
  8. Oct 31, 2006 #7


    User Avatar
    Homework Helper

    Again, read BerryBoy's post. As stated, it's a dynamics problem. You have to calculate changes of kinetic energy, which equal the work done. Then plug that work into the equation for power, which, as said, remains constant.

    P.S. Convert [km/h] to [m/s].
    Last edited: Nov 1, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook