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Math  /  Calculus

QuestionQuestion 2 (1 point) A car starts from rest and continues at a rate of v=18t2ft/sv=\frac{1}{8} t^{2} \mathrm{ft} / \mathrm{s}. Find the function that relates the distance ss the car has traveled to the time tt in seconds.

Studdy Solution

STEP 1

1. The car starts from rest, meaning initial velocity is 0.
2. The velocity of the car as a function of time is given by v(t)=18t2 v(t) = \frac{1}{8} t^2 ft/s.
3. We need to find the distance s(t) s(t) as a function of time t t .

STEP 2

1. Understand the relationship between velocity and distance.
2. Set up the integral to find the distance function.
3. Solve the integral to find the distance function.

STEP 3

The relationship between velocity and distance is given by the fact that velocity is the derivative of distance with respect to time: v(t)=dsdt v(t) = \frac{ds}{dt} .

STEP 4

To find the distance function s(t) s(t) , set up the integral of the velocity function: s(t)=∫v(t) dt=∫18t2 dt s(t) = \int v(t) \, dt = \int \frac{1}{8} t^2 \, dt

STEP 5

Solve the integral: s(t)=∫18t2 dt=18∫t2 dt s(t) = \int \frac{1}{8} t^2 \, dt = \frac{1}{8} \int t^2 \, dt
s(t)=18(t33)+C s(t) = \frac{1}{8} \left( \frac{t^3}{3} \right) + C
s(t)=124t3+C s(t) = \frac{1}{24} t^3 + C
Since the car starts from rest, the initial distance s(0)=0 s(0) = 0 . Therefore, C=0 C = 0 .
The distance function is: s(t)=124t3 s(t) = \frac{1}{24} t^3
The function that relates the distance s s to the time t t is:
s(t)=124t3 \boxed{s(t) = \frac{1}{24} t^3}

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