<u>Answer</u>
The acceleration is
to the nearest tenth
<u>Explanation</u>
Since the car was travelling at before it starts to decelerate, the initial velocity is
.
The final velocity is , because the car came to a stop.
The time taken is .
Using the Newton's equation of linear motion,
, we find the acceleration by substituting the known values.
This implies that,
This gives us,
We divide both sides by 15 to get,
or
Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
a = -1.47 m/ (a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/ (since same friction force is applied)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m