Asked by John
A police car traveling at 94.5 km/h is traveling west, chasing a motorist traveling at 77.2 km/h.
(a) What is the velocity of the motorist relative to the police car?
The magnitude and its direction
(b) What is the velocity of the police car relative to the motorist?
The magnitude and its direction
(c) If they are originally 250 m apart, in what time interval will the police car overtake the motorist?
A full working would be appreciated! thanks!
(a) What is the velocity of the motorist relative to the police car?
The magnitude and its direction
(b) What is the velocity of the police car relative to the motorist?
The magnitude and its direction
(c) If they are originally 250 m apart, in what time interval will the police car overtake the motorist?
A full working would be appreciated! thanks!
Answers
Answered by
Henry
a. Vm = 77.2 - 94.5 = -17.3 km/h.
b. Vp=94.5 - 77.2=17.3 km/h=Velocity of policeman.
c. Vm*t=X km=Dist traveled bymotorist.
Eq1: 77.2t = X.
Vp*t = (X+250)=Dist. traveled by police.
Eq2: 94.5t = X+0.250)km.
In Eq2, substitute 77.2t for x:
94.5t = 77.2t + 0.2525
94.5t - 77.2t = 0.25
17.3t = 0.25
t = 0.01453 h. = 0.87 min = 52.3 s.
b. Vp=94.5 - 77.2=17.3 km/h=Velocity of policeman.
c. Vm*t=X km=Dist traveled bymotorist.
Eq1: 77.2t = X.
Vp*t = (X+250)=Dist. traveled by police.
Eq2: 94.5t = X+0.250)km.
In Eq2, substitute 77.2t for x:
94.5t = 77.2t + 0.2525
94.5t - 77.2t = 0.25
17.3t = 0.25
t = 0.01453 h. = 0.87 min = 52.3 s.
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