Question

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



a.) Determine the position function t greater than or equal to 0.

b.)How far does the racer travel in 5 s?

c.) At this rate, how long will it take the racer to travel 1/3 mi?

d.) How long will it take the racer to travel 300ft?

e.) How far has the racer traveled when it reaches the speed of 178ft/s?

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!

Answer 2

To solve the problem we must know about the concept of Acceleration.

What is acceleration?

Acceleration is defined as the rate of change of velocity of an object with respect to time.

$a = \dfrac{dv}{dt}$

What is velocity?

Velocity is defined as the rate of change of position of an object with respect to time.

$v=\dfrac{dx}{dt}$

What is the velocity function of the racer?

As we know that acceleration is written as,

$a = \dfrac{dv}{dt}\\\\\int dv =\int a\ dt$

substitute the value a(t) = 66,

$\int dv =\int (66)\ dt$

$v = 66t + c$

As the condition given v(0) = 0,

$v = 66t + c\\\\0 = 66(0)+c\\\\c = 0$

Therefore, the velocity of the racer can be written as v=66t.

What is function for the position of the racer?

The velocity is written as,

$v=\dfrac{dx}{dt}\\\\\int dx= \int v\ dt \\\\\int dx= \int (66t)\ dt\\\\x = 33t^2+c$

Substitute the given value x(0) = 0

$0 = 33(0)^2+c\\\\c=0$

Thud, the position of the racer can be given by the function s = 33t².

A.) The position function t is greater than or equal to 0.

The position function t greater than or equal to 0 can be given by the function s = 33t².

B.) Distance traveled by the racer in 5 seconds.

To find the distance traveled by the racer in 5 seconds substitute the value of t in the position function,

s = 33(5)²

s = 825 m

C.) Time is taken by the racer to travel 1/3 mi.

We already have the position function,

s = 1/3 mi = 536.448 m

s = 33t²

536.448 = 33t²

t = 4.0319 second

D.) Time is taken by the racer to travel 300 ft

We already have the position function,

s = 300 ft = 91.44 m

s = 33t²

91.44 = 33t²

t = 1.6646 second

E.) Distance traveled by the race when it reaches a speed of 178 ft/s.

178 ft/s = 54.2544 m/s

v = 66t

54.2544 = 66t

t = 0.822 sec

Distance traveled in t = 0.8220 sec,

s = 33t²

s = 33(0.822)²

s = 22.2975 m

Learn more about Acceleration:

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