Bike race Theo and Sasha start at the same place on a straight...

Question

Bike race Theo and Sasha start at the same place on a straight road, riding bikes with the following velocities (measured in mi/hr). Assume t is measured in hours.

Theo: vT(t)=10, for t≥0

Sasha: vS(t)=15t, for 0≤t≤1, and vS(t)=15, for t>1

c. If the riders ride for 2 hr, who rides farther? Interpret your answer geometrically using the graphs of part (a).

Accepted Answer

Welcome back, everyone. Two cars, car A and B, start together and travel along a straight highway. Their velocities in kilometers per hour are Car A, the subscript A of T equals 20, or greater than or equals 0, and car B. The subscript B of T equals 30 T for T between 0 and 0.5 inclusive, and V subscript B of T is equal to 15 for T greater than 0.5, where T is in hours. After 1.5 hours, which car has traveled farther. For this problem, let's recall that distance can be identified by integrating the absolute value of the velocity function with respect to time. We're going to integrate between two time values, T1 and T2. So we can begin by finding distance traveled by car A. We have our velocity function or any time greater than or equals to zero. Our upper limit of integration is going to be 1.5 because we're looking at the time interval of 1.5. So let's integrate from 0 up to 1.5. The absolute value of 20, that's our velocity function, right? DT. We can drop the absolute value because the absolute value of 20 is just 20. We can factor out the constant and integrate DT, and the integral of DT is simply T. So we get 20 T from 0 up to 1.5. And now applying the evaluation theorem, we get 20 in C 1.5 minus 0. This is equal to 30 kilometers. That's the distance traveled by car A. Now for car B, The solution is going to be a bit more complex because now we have two velocity functions. First of all, we have our velocity function between 0 and 0.5, so we're going to integrate from 0 up to 0.5. We're going to take the absolute value of 30 CDC. Once again, we can drop the absolute value because 30T is non-negative between 0 and 0.5. And then we're going to add the second segment from 0.5 up to 1.5, right? Our final time value. And for this time interval, our velocity function is 15. We have constant velocity, once again, no necessity to add the absolute value sign, right? Integrating 30T, we get 30 T square divided by 2 according to the power rule. We went to evaluate this from 0 up to 0.5. And integrating 15 DT we get 15C, and we want to evaluate from 0.5 up to 1.5. Now, let's go ahead and simplify, 30 divided by 2 is 15, so we can just write 15 T squad for the first term. We get 15 in Cs. 0.5 squared minus 0 squad and we want to add 15 I see 1.5 minus 0.5. Performing the calculations, we get 3.75 plus 15, which is 18.75 kilometers. So now comparing the two distances, we have 30 kilometers traveled by car A and 18.75 kilometers traveled by B. So which one? Covers a larger distance. Well, that would be car A, right? So our answer would be car A. That's our final answer and thank you for watching.

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