Question

A drag racer accelerates at a(t) = 66 ft/s^2. Assume that v(0) = 0 and s(0) = 0.

Answer 1

So lets write down what we have a(t) = 66 and v(0) = 0 and s(0) = 0

a) Determine the position function.

To do this we have to integrate our acceleration function twice, or we have to integrate the acceleration function to get our velocity function and then integrate that to get our position function.

So:

$\int\limits {a(t)} \, dt = \int\limits {66} \, dt$

= 66t + c

This means that v(t) = 66t + c

We know that v(0) = 0, so:

v(0) = 66(0) + c

c = 0

So v(t) = 66t (This will be helpful for us later)

Now we have to integrate again.

$\int\limits {66t} \, dt$

$= 33t^2 + c$

*note that both of these integrands are done with the reverse power rule for integration*

So we can say that $s(t) = 33t^2 + c$

But we know that s(0) = 0

So

s(0) = 33(0)^2 + c

c = 0

So... s(t) = 33t^2

b) Now we can answer this question using our position function!

All we have to do is plug in t=5 for s(t)

So...

s(5) = 33(5)^2

s(5) = 825 ft

c) So this is essentially the same problem as b except now we are solving for the time instead of the distance.

We know that in 1 mile there are 5280 ft

So in 1/3 a mile there are 5280(1/3) ft or 1760 ft

Now we can set 1760 equal to s(t) and solve for t

1760 = 33t^2

t^2 = 53.33

$t = \sqrt{53.33}$

t = 7.30s (we only consider positive answers for time)

d) This is the same question as c just a different distance so:

Setting 300 for s(t)

300 = 33t^2

t^2 = 9.091

$t = \sqrt{9.091}$

t = 3.015s

e) So for this question we have to approach it in terms of the velocity function not the position function. Then we will solve for the time it took to travel with that velocity and then plug that time value into the position function.

So:

v(t) = 66t

We know that in this case v(t) = 178 ft/s

So: 178 = 66t

t = 2.6969 // t = 2.7s

Now we can use this time in our position function to solve for the distance traveled.

s(t) = 33t^2

s(2.7) = 33(2.7)^2

s(2.7) - 240.57 ft

Hope this helped!