Mr. Barrow scored his high school students' essays on a scale of 0 to 10. He recorded data about the scores in the table below.
2 answers:
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Answer:
(a) 2.29 km/h
(b) 9 km/h
Step-by-step explanation:
For part (a) you have to apply<em> the average speed formula</em>, which is defined by:
where d is the total distance traveled and t is the total time needed.
km/h
For part (b) you have to calculate the running time (T) , which is the total time of the race minus the nap time:
The nap time in hours is:
90/60 = 1.5 h (because there are 60 minutes in one hour)
The running time is:
T= 1.75 - 1.5 = 0.25 h
Let t1 represent the time before the nap and t2 the time after the nap:
t1+t2 = T
t1+t2 = 0.25
You have to apply the formula d=vt before and after the nap:
-Before the nap, the distance traveled was 0.50 km
0.50 = v1t1
-Afer the nap, the distance traveled was 3.50 km
3.50=v2t2
But v2=2v1 (because after the nap the rabbit runs twice as fast)
You have to solve the system of equations:
t1=0.25-t2 (I)
v1t1=0.50 (II)
2v1t2=3.50 (III)
Replacing (I) in (II)
v1(0.25-t2)=0.50
Applying distributive property and solving:
0.25v1-v1t2=.050
For (III) you have that v1t2=3.50/2=1.75. Hence:
0.25v1-1.75=0.50
Solving for v1:
v1 = 9 km/h