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Calculus/linear motion

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Expert: - 2/18/2009


Question
a sled is initially moving at a rate of 44 ft/sec. it decelerates to 32 ft/sec over a distance of 114 ft at an unknown constant rate. it continues to decelerate at the same rate until it comes to a full stop.
a) how long does it take to reduce the speed to 32 ft/sec?
b) what is the acceleration of the sled?
c) how long does it take before the sled comes to a complete stop?
d) how many feet does the sled travel before it comes to a stop?

Answer
We know that the equation is v(t) = 44 + (32-44)t/N for some t
with N as the amount of time to get there.
Note that at when t is 0, the velocity is 44.
Note that when t is N, t/N is 1, and you get v(t) is 44+32-44=32.

The next thing to note is that 32-44=-12, so the quation is
v(t) = 44 - 12t/N.

Now d(t) { distance travelled } is the integral of velocity.
⌠N
⌡0 44 - 12t/N dt = 44t - 6tē/N evaluated from (0 to N).

The value at t=0 is 0.  The value at t=N is
44N - 6Nē/N = 44N-6N = 38N.

Since we know that 38N = 114, we also know that N=3.

This would make v(t) = 44 - 12t/3 = 44 - 4t.

By the way, the answer to (a) was t=3.

The answer to (b) is a=-4.

To answer (c), find out what value of t would make the velocity 0.
Note that v(t) = 44 - 4 and you want to solve for t when v(t)=0.

To answer (d), use d(t).  The function d(t) can be found above to
be 44t - 2tē.

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