2. A sports car is stopped at a light. At t = 0 the light changes and the sports car accelerates at a
constant 2.0 m/s2. At t = (10/3) s a station wagon traveling at a constant 15 m/s in the same
direction passes the stop light. When does the station wagon catch up to the sports car?
Answers
Answer:
This one is pretty straightforward. You know that at
t
=
0
and
x
=
0
, the car starts moving with a constant acceleration of
2.1 m/s
2
.
At the same time, a truck moving at a speed of
9.2 m/s
passes the car.
The idea here is that the distance they travel until the car overtakes the truck is the same for both the car, and the truck.
The same can be said for the time that passes until the two meet again.
So, if the car takes a time
t
to reach the truck, and both travel a distance
d
before that happens, then you an say that
d
=
v
⋅
t
→
for the truck
and
d
=
v
0
=
0
⋅
t
+
1
2
⋅
a
⋅
t
2
→
for the car
This will get you
v
⋅
t
=
1
2
⋅
a
⋅
t
2
⇒
t
=
2
⋅
v
a
t
=
2
⋅
9.2
m
s
2.1
m
s
2
=
8.76 s
This means that the distance is
d
=
v
⋅
t
=
9.2
m
s
⋅
8.76
s
=
80.6 m
The speed of the car is
v
car
=
a
⋅
t
=
2.1
m
s
2
⋅
8.76
s
=
18.4 m/s