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?
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.
To find the speed of the police officer when he reaches the motorist, we need to determine the time it takes for the officer to catch up with the motorist.
Let's denote the time it takes for the officer to catch up as T. At time T, the positions of the officer and the motorist will be the same.
Now let's determine the position of the motorist as a function of time. Since the motorist is traveling at a constant speed of 60 m/s, we can write the position of the motorist as:
x_motorist = 60t
Next, let's determine the position of the officer as a function of time. The officer starts from rest at time t = 0, and accelerates according to the formula a(t) = bt. We can integrate the acceleration function to find the velocity function:
v_officer = ∫(a(t)) dt = ∫(bt) dt = (b/2)t^2 + C1
Now, since the officer starts from rest, at time t = 0, the officer's velocity is 0. Therefore, C1 = 0, and the velocity function becomes:
v_officer = (b/2)t^2
Similarly, we can integrate the velocity function to find the position function:
x_officer = ∫(v_officer) dt = ∫((b/2)t^2) dt = (b/6) t^3 + C2
Again, since the officer starts from rest, at time t = 0, the officer's position is 0. Therefore, C2 = 0, and the position function becomes:
x_officer = (b/6) t^3
Now, as mentioned earlier, at time T, the positions of the officer and the motorist are equal. So we can equate the position functions:
x_motorist = x_officer
Substituting the expressions for x_motorist and x_officer:
60t = (b/6) t^3
Simplifying the equation:
10t = (b/6) t^3
Dividing both sides of the equation by t:
10 = (b/6) t^2
Simplifying further:
t^2 = (60/3b)
t = √(60/3b)
Now, since we want to find the speed of the officer when he reaches the motorist, we can substitute this value of t back into the velocity function for the officer:
v_officer = (b/2)t^2
v_officer = (b/2)(√(60/3b))^2
Simplifying further:
v_officer = (b/2)(60/3b)
v_officer = 10 m/s
Therefore, the speed of the police officer when he reaches the motorist is 10 m/s.