Physics 1307 Quiz Chapters 23 Solution
Physics 1307 Quiz Chapters 23 Solution Problem A speeding motorist zooms through a 50 km/h zone at 75 km/h without noticing a stationary police car. The police officer heads after the speeder, accelerating at 2.5 m/s². When the officer catches up to the speeder, how far down the road are they? Draw a diagram of the situation and plot x vs t diagram for both the speeder and the officer and give your answer in m. x
v=75 km/h
xc
v=0 a=2.5 m/s2
x tc
t
Using the drawing we define the origin (x=0) as the point where the police car is at time t =0. At this moment both cars have the same position. Our job is to find the second point of intersection, thus we need to find xc , which implies that we also need tc. Since the position of the speeding car is given by the equation of motion with constant velocity we have: 1000 m x s =x 0vt=75
km 1 km h 3600 s 1h
x s=21
1/2
m s
t
t
similarly for the police car : 1 x p =x 0v p t a t 2 2 1 x p= 2.5 m /s 2 t 2 2 when the police car catches up to the speeder both positions are equal, therefore: x p =x s 1 2.5 m /s 2 t c2 = 21 m/s t c 2 This is a quadratic equation with solutions, t c =0 s t c=
2×21m/ s 2.5m /s
2
=16.8 s
we choose the second solution since the first one corresponds to the beginning of the chase. The distance can be found using any equation , the easier one is that of the speeder: x c =21m/ s t c =353 m Extra credit If the v vs t diagram of an object is given by the diagram below, draw the corresponding a vs t diagram v
a t1
t2
t
2/2
t1
t2
t
v=75 km/h
xc
v=0 a=2.5 m/s2
x tc
t
Using the drawing we define the origin (x=0) as the point where the police car is at time t =0. At this moment both cars have the same position. Our job is to find the second point of intersection, thus we need to find xc , which implies that we also need tc. Since the position of the speeding car is given by the equation of motion with constant velocity we have: 1000 m x s =x 0vt=75
km 1 km h 3600 s 1h
x s=21
1/2
m s
t
t
similarly for the police car : 1 x p =x 0v p t a t 2 2 1 x p= 2.5 m /s 2 t 2 2 when the police car catches up to the speeder both positions are equal, therefore: x p =x s 1 2.5 m /s 2 t c2 = 21 m/s t c 2 This is a quadratic equation with solutions, t c =0 s t c=
2×21m/ s 2.5m /s
2
=16.8 s
we choose the second solution since the first one corresponds to the beginning of the chase. The distance can be found using any equation , the easier one is that of the speeder: x c =21m/ s t c =353 m Extra credit If the v vs t diagram of an object is given by the diagram below, draw the corresponding a vs t diagram v
a t1
t2
t
2/2
t1
t2
t
