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≥0Sasha: vS(t)=15t, for 0≤t≤1, and vS(t)=15, for t\u003e1f. Suppose Sasha gives Theo a head start of 0.2 hr and the riders ride for 20 mi. Who wins the race?

Accepted Answer

First, understand the problem setup: Theo starts riding at time t = 0 with a constant velocity $$v_T(t) = 10 mi/hr. $$Sasha starts after a head start of 0.2 hours, so Sasha's motion begins at $$t = 0.2 hr$$, and Sasha's velocity is piecewise defined as $$v_S(t) = 15t$$ for $$0 \leq t \leq 1$$ and $$v_S(t) = 15$$ for $$t \u003e 1 ($$where $$t is $$measured from Sasha's start time). Calculate the time it takes for Theo to complete the 20-mile race. Since Theo's velocity is constant, use the formula for time: $$t_T = \frac{	ext{distance}}{	ext{velocity}} = \frac{20}{10}. $$This gives the total time Theo rides from $$t=0$$ until finishing. For Sasha, note that Sasha starts at $$t=0.2 hr ($$relative to the original clock). Define Sasha's travel time as $$t_S ($$time after Sasha starts). We need to find $$t_S$$ such that the total distance Sasha covers equals 20 miles. Since Sasha's velocity changes at $$t_S=1$$ hour, break the problem into two parts if needed: from $$0 to$$ $$1$$ hour and from $$1$$ hour onward. Set up the distance function for Sasha by integrating the velocity over time. For $$0 \leq t_S \leq 1$$, the distance is $$d_S(t_S) = \$$int_0$$^{t_S} 15u \, du. $$For $$t_S \u003e 1$$, the distance is $$d_S(t_S) = \$$int_0$$^1 15u \, du + \$$int_1$$^{t_S} 15 \, du. $$Solve for $$t_S$$ such that $$d_S(t_S) = 20$$ miles. Compare the finishing times: Theo finishes at $$t = t_T$$, and Sasha finishes at $$t = 0.2 + t_S ($$since Sasha started 0.2 hours later). The rider with the smaller finishing time wins the race.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Velocity and Speed Functions

Distance as the Integral of Velocity

Time Delay and Relative Start Times