Question

2.3: constant rate of change and linear functions (homework)
when running a 100 - meters race salvador reaches his maximum speed when he is 15 meters from the starting line and 3 seconds have elapsed since the start of the race. salvador maintains this maximum (constant) speed for the rest of the race. if salvador is 51 meters from the starting line 7 seconds after the start of the race.
a. what is salvadors max speed?
meters per second preview
b. if salvador runs for an additional 2 seconds after reaching his max speed.
i. how far will salvador travel during those 2 seconds?
meters preview
ii. what is salvadors distance from the starting line 5 seconds after the start of the race?
meters preview
c. what is salvadors distance from the starting line 4.5 seconds after the start of the race?
meters preview
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when running a 100 - meter race keyon reaches his maximum speed 3 seconds after the race started when he is 16

Explanation:

Step1: Calculate max - speed

Salvador runs $(51 - 15)$ meters in $(7 - 3)$ seconds. Speed formula is $v=\frac{d}{t}$. So $v=\frac{51 - 15}{7 - 3}=\frac{36}{4}=9$ meters per second.

Step2: Distance in 2 seconds

Using $d = vt$, with $v = 9$ m/s and $t = 2$ s, $d=9\times2 = 18$ meters.

Step3: Distance at 5 seconds

He runs 15 meters in 3 seconds. In the next 2 seconds (from 3 - 5 seconds) at 9 m/s, he runs $9\times2$ meters. Total distance is $15+9\times2=15 + 18=33$ meters.

Step4: Distance at 4.5 seconds

From 3 - 4.5 seconds is 1.5 seconds. Distance in this time is $9\times1.5 = 13.5$ meters. Total distance is $15+13.5=28.5$ meters.

Answer:

a. 9
b. i. 18
ii. 33
c. 28.5