Math  /  Calculus

QuestionThis is Section 4.4 Problem 54: PREVIOUS ANSWER
Two cars enter a freeway at the same time. The velocities, measured by miles per minute, of Car A and CarBt\mathrm{Car} \mathrm{B} t minutes after entering the freeway are given by vA(t)=0.4+0.2t0.02t2,0t5vB(t)=0.1+0.3t0.02t2,0t5\begin{array}{l} v_{A}(t)=0.4+0.2 t-0.02 t^{2}, 0 \leq t \leq 5 \\ v_{B}(t)=0.1+0.3 t-0.02 t^{2}, 0 \leq t \leq 5 \end{array}
At the end of 5 minutes of driving, Car A has traveled \square miles, and Car B has traveled \square miles. Hence CarA \checkmark \checkmark is ahead of Car B \checkmark \checkmark by \square Hint: Follow Example 5. Additional Materials ebook

Studdy Solution

STEP 1

What is this asking? We're given the speeds of two cars entering a freeway at the same time, and we need to figure out how far each car has traveled after 5 minutes and which car is ahead. Watch out! We're given *velocity* (speed), not distance.
Don't forget to calculate the distance traveled from the velocity functions!

STEP 2

1. Calculate the distance traveled by Car A.
2. Calculate the distance traveled by Car B.
3. Compare the distances.

STEP 3

Alright, so we're given Car A's velocity vA(t)v_A(t) as a function of time tt.
To find the distance traveled, we need to find the *area under the velocity curve*.
This means we need to **integrate** the velocity function from the **initial time** t=0t = 0 to t=5t = 5 minutes.

STEP 4

Let's **set up the integral** for Car A: 05(0.4+0.2t0.02t2)dt \int_0^5 (0.4 + 0.2t - 0.02t^2) dt

STEP 5

Now, let's **integrate** term by term: [0.4t+0.2t220.02t33]05 \left[ 0.4t + \frac{0.2t^2}{2} - \frac{0.02t^3}{3} \right]_0^5 [0.4t+0.1t20.023t3]05 \left[ 0.4t + 0.1t^2 - \frac{0.02}{3}t^3 \right]_0^5

STEP 6

**Plug in** the **upper limit** of integration (t=5t = 5): 0.4(5)+0.1(52)0.023(53) 0.4(5) + 0.1(5^2) - \frac{0.02}{3}(5^3) 2+2.50.023(125) 2 + 2.5 - \frac{0.02}{3}(125) 4.52.53 4.5 - \frac{2.5}{3} 4.556 4.5 - \frac{5}{6}

STEP 7

To subtract the fractions, let's **convert** 4.54.5 to a fraction with a **denominator** of 6: 27656=226=113 \frac{27}{6} - \frac{5}{6} = \frac{22}{6} = \frac{11}{3} So, Car A traveled 113\frac{11}{3} miles.

STEP 8

We'll do the same thing for Car B.
Its velocity is vB(t)=0.1+0.3t0.02t2v_B(t) = 0.1 + 0.3t - 0.02t^2.
We need to **integrate** this from t=0t = 0 to t=5t = 5.

STEP 9

**Set up the integral** for Car B: 05(0.1+0.3t0.02t2)dt \int_0^5 (0.1 + 0.3t - 0.02t^2) dt

STEP 10

**Integrate** term by term: [0.1t+0.3t220.02t33]05 \left[ 0.1t + \frac{0.3t^2}{2} - \frac{0.02t^3}{3} \right]_0^5 [0.1t+0.15t20.023t3]05 \left[ 0.1t + 0.15t^2 - \frac{0.02}{3}t^3 \right]_0^5

STEP 11

**Plug in** t=5t = 5: 0.1(5)+0.15(52)0.023(53) 0.1(5) + 0.15(5^2) - \frac{0.02}{3}(5^3) 0.5+3.752.53 0.5 + 3.75 - \frac{2.5}{3} 4.2556 4.25 - \frac{5}{6} 51121012=4112 \frac{51}{12} - \frac{10}{12} = \frac{41}{12} So, Car B traveled 4112\frac{41}{12} miles.

STEP 12

Car A traveled 113\frac{11}{3} miles, which is 4412\frac{44}{12} miles.
Car B traveled 4112\frac{41}{12} miles.

STEP 13

Since 4412>4112\frac{44}{12} > \frac{41}{12}, Car A is ahead of Car B by 44124112=312=14\frac{44}{12} - \frac{41}{12} = \frac{3}{12} = \frac{1}{4} miles.

STEP 14

After 5 minutes, Car A has traveled 113\frac{11}{3} miles, and Car B has traveled 4112\frac{41}{12} miles.
Car A is ahead of Car B by 14\frac{1}{4} miles.

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