Question

a sled accelerates down a hill and then slows down after it reaches a flat portion of ground. the speed of the sled ( v(t) ) (in ft/sec) at a time ( t ) (in sec) after movement begins can be approximated by: v(t) = \begin{cases} 2t & \text{for } 0 leq t leq 20 \\ dfrac{80}{t - 19} & \text{for } 20 < t leq 70 end{cases} determine the speed of the sled after 10 sec, 20 sec, 30 sec, and 70 sec. round to 1 decimal place if necessary. \\ part 1 of 4 \\ the speed of the sled is \\( square \\) ft/sec after 10 sec. \\ part 2 of 4 \\ the speed of the sled is \\( square \\) ft/sec after 20 sec.

Explanation:

Step1: Evaluate at t=10 sec

Since $0 \leq 10 \leq 20$, use $v(t)=2t$.
$v(10)=2\times10=20$

Step2: Evaluate at t=20 sec

Since $0 \leq 20 \leq 20$, use $v(t)=2t$.
$v(20)=2\times20=40$

Step3: Evaluate at t=30 sec

Since $20 < 30 \leq 70$, use $v(t)=\frac{50}{t-19}$.
$v(30)=\frac{50}{30-19}=\frac{50}{11}\approx4.5$

Step4: Evaluate at t=70 sec

Since $20 < 70 \leq 70$, use $v(t)=\frac{50}{t-19}$.
$v(70)=\frac{50}{70-19}=\frac{50}{51}\approx1.0$

Answer:

• The speed of the sled is 20 ft/sec after 10 sec.
• The speed of the sled is 40 ft/sec after 20 sec.
• The speed of the sled is 4.5 ft/sec after 30 sec.
• The speed of the sled is 1.0 ft/sec after 70 sec.