Asked by Jason
A motorist drives along a straight road at a constant speed of 60 m/s. At time t = 0 she passes a parked motorcycle
police o±cer. The o±cer takes o® after her at t = 0 and accelerates according to the formula a(t) = bt, where t is the
time and b is a positive constant. What is the speed of the police o±cer when he reaches the motorist?
police o±cer. The o±cer takes o® after her at t = 0 and accelerates according to the formula a(t) = bt, where t is the
time and b is a positive constant. What is the speed of the police o±cer when he reaches the motorist?
Answers
Answered by
bobpursley
Well, they travel the same distance in the same time.
The officers velocity is INT a(t)dt= INT btdt= 1/2 b t^2
so the officers distance traveled is
INT 1/2 b t^2 dt= 1/6 b t^3
set the distances equal
1/6 b t^3= 60t
solve for t.
t^2= 360/b
t= sqrt(360/b)
now find the velocity of the officer at that point:
Vf= 1/2 b t^2= 1/2 b * 360/b=180m/s
check my work.
The officers velocity is INT a(t)dt= INT btdt= 1/2 b t^2
so the officers distance traveled is
INT 1/2 b t^2 dt= 1/6 b t^3
set the distances equal
1/6 b t^3= 60t
solve for t.
t^2= 360/b
t= sqrt(360/b)
now find the velocity of the officer at that point:
Vf= 1/2 b t^2= 1/2 b * 360/b=180m/s
check my work.
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