Acy list rides down a long straightr ad at a velocity in a through c. a. How far

Question

Acy list rides down a long straightr ad at a velocity in a through c. a. How far does the cyclist travel in the first 4 min? given by t=400 20t, for 0 sis 10. Complete parts min b. How far does the cyclist travel in the first 6 min? c. How far has the cyclist traveled when his velocity is 290 ? min a. The cyclist travels 1440 m in the first 4 min. b. The oyclist travels 2040 m in the first 6 min min (Round to two decimal places as needed.) (o) More Enter your answer in the answer box and then elick Check Answer Clear All All parts showing

Explanation / Answer

We have given velocity v(t)=400-20t,0<=t<=10

a) the cyclist travel in the first 4 minutes is displacement s(4)= integration of (0 to 4)v(t)dt

integration of (0 to 4)v(t)dt=integration of (0 to 4)(400-20t)dt

=[400t-20t2/2] from 0 to 4

=[400t-10t2] from 0 to 4

=[400*4-10(4)2]-[0-0]

=1600-160

=1440

The cyclist travels 1440 m in the first 4 minutes.

b) the cyclist travel in the first 6 minutes is displacement s(6)= integration of (0 to 6)v(t)dt

integration of (0 to 6)v(t)dt=integration of (0 to 6)(400-20t)dt

=[400t-20t2/2] from 0 to 6

=[400t-10t2] from 0 to 6

=[400*6-10(6)2]-[0-0]

=2400-360

=2040

The cyclist travels 2040 m in the first 6 minutes.

c) We have given v(t)=290 m/min

when v(t)=400-20t=290

-20t=290-400=-110

t=110/20=11/2

so the displacement s(11/2)=integration of (0 to 11/2) v(t)dt

=integration of (0 to 11/2) (400-20t)dt

=[400t-10t2] from 0 to 11/2

=[400*(11/2)-10*(11/2)^2]-[0]

=2200-(10*121)/4

=(4400-5*121)/2

=1897.5

When the cyclist's velocity is 290 m/min he has traveled 1897.5 m

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