Question

a sled accelerates down a hill and then slows down after it reaches a flat portion of ground. the speed of the sled ( s(t) ) (in ft/sec) at a time ( t ) (in movement begins can be approximated by: s(t) = \begin{cases} 2t & \text{for } 0 leq t leq 20 \\ dfrac{50}{t - 19} & \text{for } 20 < t leq 70 end{cases} determine the speed of the sled after 10 sec, 20 sec, 50 sec, and 70 sec. round to 1 decimal place if necessary. part 1 of 4

Explanation:

Step1: Evaluate at t=10 sec

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

Step2: Evaluate at t=20 sec

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

Step3: Evaluate at t=50 sec

Since $20 < 50 \leq 70$, use $s(t)=\frac{50}{t-19}$.
$s(50)=\frac{50}{50-19}=\frac{50}{31}\approx1.6$

Step4: Evaluate at t=70 sec

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

Answer:

• After 10 sec: 20.0 ft/sec
• After 20 sec: 40.0 ft/sec
• After 50 sec: 1.6 ft/sec
• After 70 sec: 1.0 ft/sec