Finding acceleration

  • Thread starter mistabry
  • Start date
  • #1
12
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Homework Statement


You are driving along a highway at 80.0 km/hr in car that is 6.0 m long. Exactly 10 m ahead of you is a truck of length 12.0 m driving at 60.0 km/hr.

a. If you wish to pass the truck so that the back of your car just clears the front of the truck in 3.0 s, how fast must you accelerate?
b. What will your speed be when you pass the front end of the truck?


Homework Equations


I thought I could use Xf-Xi = Vixt + 1/2a(t)^2, but it didn't give me the correct answer.


The Attempt at a Solution


Plugged in Xf=28km Vix=80km/h and t=3, but I couldn't get the correct answer. I didn't attempt B because I couldn't find A >.<
 

Answers and Replies

  • #2
If you are associating coordinate X with moving truck (truck's beginning or end), then Vi should be cars speed relative to the truck, not 80 [km/h]. Distances are given in meters, time in seconds, speed in kilometers per hour. So first make all units to be the same. Don't put different types of units into one equation.
 
  • #3
6
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The best way would be to define your relative speed first which is 20 kmh. So you are already catching up with him. The distance to travel would be 28.0 m just like you said. 20 kmh = 5.55 m/s. Because you are already driving at a certain speed and you are accelerating you have to combine some factors. Theoraticly the surface between the V(t) and the x-axis is equal to the distance you travel. v(t) = at + 5.55
The intergral of this function describing your surface is equal to: s(t) = 1/2at2 + 20t

if you fill in the variables: 28 = 0.5 * a * 9 + 3 * 5.55

And I think you can do the rest yourself.
In the end you'll have to use the following function: s = 1/2 at2 + vt
 
  • #4
12
0
Thank you Aviation. I got the answers right with your method with a. 2.522 m/s^2 and b. 107.24 km/hr
 

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