I know I've asked quite a few questions, but I really am making an effort to get as far as I can. The engineer of a train moving along a level track with a velocity of 42.0 m/s sights a freight train at a distance of d ahead of him on the same track moving in the same direction with a velocity of 18.0 m/s. He applies the brakes, giving his train a constant acceleration of -1.4 m/s^2. What is the minium distance d such that there is no collision?(adsbygoogle = window.adsbygoogle || []).push({});

Here's what I came up with. I used the formula d/delta x = 42.0t+1/2(-1.4)t^2. I derived the formula giving me 42+(-1.4)=18. Solved for t getting 17.143. Plugged t back into the original formula I used and ended up getting 514.288. That isn't the correct distance..

Here's another question..

Let's say that you are driving a car that accelerates according to a=B/v where B= 130.0m^2/s^3 is a parameter that is related to the ratio of your car's power to its weight, and c is your car's speed, in m/s. Assume that you are initially traveling with a speed of 11.5 m/s. At t=0 you step on the gas pedal. The car performs a constant-power acceleration until you reach a speed of 23.5 m/s. What is the time interval needed to make this change of speed.

I solved for a by plugging in the values given for B and v, so a=130.0m^2/s^3/11.5 m/s. I then got a= 11.3m/s^2. From there I went to 11.5+11.3t=23.5. Solved for t and got 1.062 s. 1.062s isn't the correct answer. Was my approach to this problem incorrect?

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# Homework Help: Minimum distance

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