Accelerated Linear Motion

  • #1
Hey, id greatly appreciate any urgent help you can give me with this problem

Two cars A and B, each 5m in length, travel with constant velocity 20 m/s along a straight level road. The front of car A is 15m directly behind the rear of car B. Immediately on reaching a point P, each car decelerates at 4m/s^2.
1) Show that A collides with B
2) At what distance from P does the collision occur?
3) Show the motion of both cars on the same speed-time graph.


Thanks in advance
 

Answers and Replies

  • #2
Please! Im seriously desperate here!
 
  • #3
HallsofIvy
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I know! You also posted this on the "mathematics" forum.

A crucial point here is that since each car begins decelerating when it "reaches point P", car A does not begin decelerating until AFTER car B does. In fact, since the front of car A is 20 m behind the front of car B (I assume they "reach point P" when the front of the car does) and car A is moving at 20 m/s, car A will start decelerating exactly 1 second after car B does.
With constant deceleration of 4 m/s2 and taking t= 0 when the front of car B passes point P, vB(t) (the speed of car B at time t) is 20- 4t and so xB(t) (the distance from point P to the front of car B at time t) is 20t- 2t2.

A has the same initial speed and the same deceleration but starts a second later: vA= 20- 4(t-1) and
xA(t)= 20(t-1)-2(t-1)2 (notice the clever use of (t-1) as the variable- I could have said the anti-derivative of 20 is 20t but then I would have to add -20 to make xA= 0 when t= 1 rather than 0.)

The two cars will collide when the FRONT of car A is at the same point as the BACK of car B. Since xB and xA are the distance from P to the fronts of the cars, the collision will happen when xB- xA= 5. Set up that equation and solve for t. Once you find that, substitute t into the equations for xB and xA to find the distances.
 

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