Two cars are traveling along a straight line in
the same direction, the lead car at 22 m/s and
the other car at 34 m/s. At the moment the
cars are 59 m apart, the lead driver applies the
brakes, causing the car to have a deceleration
of 1.5 m/s
2
.
How long does it take for the lead car to
stop?
Answer in units of s
A. frictional force
B. gravitational force
C. normal force
D. weak nuclear force
To find the time it takes for the lead car to stop, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, since the lead car stops)
u = initial velocity (22 m/s)
a = deceleration (-1.5 m/s^2, since it is opposing the motion)
s = distance (59 m)
We want to solve for time (t), so let's rearrange the equation:
0 = (22)^2 + 2(-1.5)s
0 = 484 - 3s
3s = 484
s = 484/3
Now we can substitute the value of s into the equation:
0 = 484 - 3(59)
0 = 484 - 177
0 = 307
Since the equation does not equate to 0, it means that the lead car does not stop within the given distance of 59 m.
Therefore, the lead car does not stop.