Answer:
The magnitude of the car's acceleration as it slows during braking is 36.81 m/s²
Explanation:
From the question, the given values are as follows:
Initial velocity, u = 90 m/s
final velocity, v = 0 m/s
distance, s = 110 m
acceleration, a = ?
Using the equation of motion, v² = u² + 2as
(90)² + 2 * 110 * a = 0
8100 + 220a = 0
220a = -8100
a = -8100/220
a = -36.81 m/s²
The value for acceleration is negative showing that car is decelerating to a stop. The magnitude of the car's acceleration as it slows during braking is therefore 36.81 m/s²
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Answer:
a) velocity v = 322.5m/s
b) time t = 19.27s
Explanation:
Note that;
ads = vdv
where
a is acceleration
s is distance
v is velocity
Given;
a = 6 + 0.02s
so,
Remember that
substituting s = 2km =2000m, into equation 1
v = 322.5m/s
substituting s = 2000m into equation 2
t = 19.27s
Answer:
the impulse experienced by the passenger is 630.47 kg
Explanation:
Given;
initial velocity of the car, u = 0
final velocity of the car, v = 9.41 m/s
time of motion of the car, t = 4.24 s
mass of the passenger in the car, m = 67 kg
The impulse experienced by the passenger is calculated as;
J = ΔP = mv - mu = m(v - u)
= 67(9.41 - 0)
= 67 x 9.41
= 630.47 kg
Therefore, the impulse experienced by the passenger is 630.47 kg