Question

1. a toy car slides down a ramp and coasts to a stop after 5 sec. its velocity from t = 0 to t = 5 is modeled by v(t) = t² - 0.2t³ ft/sec. how far does it travel?

Explanation:

Step1: Recall the relationship between velocity and distance

The distance traveled by an object is the integral of its velocity function over the time interval. So, we need to compute the definite integral of \( v(t) = t^2 - 0.2t^3 \) from \( t = 0 \) to \( t = 5 \). The formula for distance \( s \) is \( s=\int_{a}^{b}v(t)dt \), here \( a = 0 \), \( b = 5 \), and \( v(t)=t^{2}-0.2t^{3} \).

Step2: Compute the antiderivative of \( v(t) \)

First, find the antiderivative of \( t^2 \) and \( - 0.2t^3 \) using the power rule for integration, which states that \( \int x^n dx=\frac{x^{n + 1}}{n+1}+C \) (for \( n
eq - 1 \)).

• For \( \int t^2 dt \), using the power rule with \( n = 2 \), we get \( \frac{t^{2 + 1}}{2+1}=\frac{t^{3}}{3} \).
• For \( \int-0.2t^{3}dt=- 0.2\int t^{3}dt \), using the power rule with \( n = 3 \), we get \( - 0.2\times\frac{t^{3+1}}{3 + 1}=-0.2\times\frac{t^{4}}{4}=-\frac{0.2}{4}t^{4}=-\frac{1}{20}t^{4} \) (since \( 0.2=\frac{1}{5} \), so \( \frac{0.2}{4}=\frac{1}{20} \)).

So the antiderivative \( F(t) \) of \( v(t) \) is \( F(t)=\frac{t^{3}}{3}-\frac{1}{20}t^{4}+C \).

Step3: Evaluate the definite integral

Using the fundamental theorem of calculus, \( \int_{a}^{b}v(t)dt=F(b)-F(a) \). Here, \( a = 0 \), \( b = 5 \), so we compute \( F(5)-F(0) \).

First, compute \( F(5) \):
\( F(5)=\frac{5^{3}}{3}-\frac{1}{20}\times5^{4}=\frac{125}{3}-\frac{1}{20}\times625 \)
\( \frac{125}{3}-\frac{625}{20}=\frac{125\times20-625\times3}{60}=\frac{2500 - 1875}{60}=\frac{625}{60}=\frac{125}{12}\approx10.4167 \)

Then, compute \( F(0) \):
\( F(0)=\frac{0^{3}}{3}-\frac{1}{20}\times0^{4}=0 - 0 = 0 \)

Now, \( \int_{0}^{5}(t^{2}-0.2t^{3})dt=F(5)-F(0)=\frac{125}{3}-\frac{625}{20}-0 \)

Simplify \( \frac{125}{3}-\frac{625}{20} \):
Find a common denominator, which is 60.
\( \frac{125\times20}{3\times20}-\frac{625\times3}{20\times3}=\frac{2500}{60}-\frac{1875}{60}=\frac{2500 - 1875}{60}=\frac{625}{60}=\frac{125}{12}\approx10.4167 \) (or as a fraction, \( \frac{125}{12} \) ft, or approximately \( 10.42 \) ft)

Answer:

The distance the toy car travels is \( \frac{125}{12}\approx10.42 \) feet (or exactly \( \frac{125}{12} \) feet).