A certain cable car in San Francisco can stop in 10 s when traveling at maximum

Question

A certain cable car in San Francisco can stop in 10 s when traveling at maximum speed. On one occasion, the driver sees a dog a distance d in front of the car and slams on the brakes instantly. The car reaches the dog 7.90 s later, and the dog jumps off the track just in time. If the car travels 3.97 m beyond the position of the dog before coming to a stop, how far was the car from the dog? Please explain why you are using which equation of motion. I want to understand how to pick the correct equations to use.

Explanation / Answer

Let us denote the decelaration that the car undergoes on hitting the brakes to be 'a'

Now, we as mentioned in the question, the car stops in 10 seconds and reaches the dog in 7.9 seconds which implies that the car will travel for another 2.1 seconds before it stops. It is further mentioned that it travels for 3.97 metres beyond the position of the dog.

So what we have for the section beyond the position of the dog is: final velocity is 0, we assume the moment it crosses dog's postion it has a velocity of u m/s. It takes 2.1 seconds to decelerate from u to 0 over a distance on 3.97 metres.

Now v = u -a*2.1; that is, U = 2.1*a [Equation 1]

Again, S = U*t - 1/2 * a * t^2 = 2.1*a*2.1 - 1/2 * a* 2.1*2.1 = 2.205a

But, S for the section beyond dog's postion = 3.97 = 2.205a; That is, a = 1.8 m/s^2 is the deceleration.

and U = 2.1*a = 3.78 m/s

Now for the section before the dog's position, we have: final velocity(at moment it crosses the dog) = 3.78 m/s

Assume the initial maximum velocity to be V. Deceleration = 1.8 m/s^2

Here again, Final velocity = Initial velocity - a*t

That is 3.78 = V - 2.1*7.9; That is, V = 20.37 m/s

Also, D = U*t - 1/2 * a * t*t

D = 20.37 * 7.9 - 2.1*7.9*7.9/2 = 160.923 - 65.5305 = 95.39 metres.

NOTE: For selection of equation of motion, you need to sort out the known values first and then choose the equation that would best fit the situation. As in the solution above, I had to use two separate equations, as I had two unknowns. It's something that you would get very easily comfortable with once you solve a decent number of problems. There is not much conceptual aspect to it, rather it's mathematical. You have to identify the knowns and unknowns and choose the equation according. For the start, I would suggest writing down the equations at one place and keep referring back to them while solving. And try to use the basic forms of the equations rather than derived forms.

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