dep 18819966 A person takes a bip, driving with a constant speed of 98.0 km/h es
ID: 1880583 • Letter: D
Question
dep 18819966 A person takes a bip, driving with a constant speed of 98.0 km/h escept for a 22.o min rest stop. The ton's average speed is 63.1 km/h. (a) How much time is spent on the trip? (b) How far does the person travel? (a) Let t, be the time the motorist travels at the constant speed of v-98.0 km/t. Since the person is not traveling during the rest stop, the total displacement for the entire trip is The time for the entire trip Auta is equal to the time ti that the motorist travels at a constant speed plus the time tg spent at the rest stop. We have The total displacement Aots is equal to the average speed of the trip v time of the trip. We have 63.1 km/h multbiplied by the total Substituting our first expression in terms of v for Abotal gives Now se can solve this equation for ty. We need to convert the given Simme spent at the rest stop from minutes to hours for t and thus we have Your response differs from the correct answer by more than 10% Double check your calculations V min Your response differs from the correct answer by more than 10%. Double check your calculations. km/h DOLLExplanation / Answer
Comment below if you have any doubt.
1.
Average Speed is given by:
V_avg = Total distance traveled/time taken
Suppose he was driving car for 't' hour, then
total time taken = t + 22 min = t + (22/60) hr = t + 11/30
Now distance covered while traveling will be
d1 = Speed*time = 98.0*t
distance covered during rest stop = 0
Total distance covered = 98t
So,
Avg speed = 63.1 km/hr
63.1 = 98t/(t + 11/30)
63.1*t + 63.1*11/30 = 98t
t = 63.1*11/(30*(98 - 63.1))
t = 0.66 hr
Total time spent on trip = t + 11/30 = 0.66 + (11/30) = 1.03 hr
Part B.
Total distance traveled will be
distance = speed*time
distance = (98 km/hr)*0.66 hr = 64.68 km
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