Answer:
-0.64525g
Explanation:
t = Time taken for the car to stop
u = Initial velocity = 95 km/h
v = Final velocity = 0 km/h
s = Displacement
a = Acceleration
Equation of motion
Converting to m/s²
g = Acceleration due to gravity = 9.81 m/s²
Dividing both the accelerations, we get
Hence, acceleration of the car is -0.64525g
Answer: The frequency heard will be f = 275.675Hz
Explanation: When an object emitting sound is moving, it occurs a phenomenon called Doppler shift or Doppler effect. What happens is that the sound gets higher when the moving object comes closer the observer and becomes lower after it passes, This change is due to the quantity of waves that passes through an area in an unit of time.
The formula to calculate the Doppler effect is as follows
f = () · f₀
f is the observed frequency;
c is the speed of sound;
Vs is velocity of the source;
f₀ is the emitted frequency of source;
Substituting and calculating,
f = · 300
f = 275.675 Hz
Thus, the frequency heard by the police officer is 275.675Hz.
Answer:
141.78 ft
Explanation:
When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.
Calculating the acceleration using one of Newton's equations of motion:
Note: The negative sign denotes deceleration.
When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2
Hence, we can find the minimum stopping distance using:
The minimum stopping distance is 141.78 ft.