Answer:
The drag coefficient of the car is 0.189
Explanation:
mass of the car = 2250 kg
Frontal area of the car = 2.35 m^2
initial speed of the car = 72 km/hr = (72 x 1000)/3600 = 20 m/s
final speed of the car = 54 km/hr = (54 x 1000)/3600 = 15 m/s
time taken by the car to slow down = 105 sec
We'll assume that the value of the drag coefficient is constant throughout the deceleration.
The car decelerates from 20 m/s to 15 m/s in 105 seconds, the deceleration is calculated from
[tex]a = \frac{v-u}{t}[/tex]
where a is the deceleration
v is the final speed of the car
u is the initial speed of the car
t is the time taken to decelerate.
imputing values, we'll have
[tex]a = \frac{15-20}{105}[/tex] = -0.0476 m/s^2 (the -ve sign indicates a deceleration, which is a negative acceleration)
we can safely ignore the -ve sign in other calculations that follows
The force (drag force) with which the air around the decelerates the car is equal to..
[tex]F_{D} = ma[/tex]
where [tex]F_{D}[/tex] is the drag force
m is the mass of the car
a is the deceleration of the car
imputing values, we'll have
[tex]F_{D} = 2250*0.0476[/tex] = 107.1 N
equation for drag force is
[tex]F_{D} = \frac{1}{2}pAC_{D} v^{2}[/tex]
where p is the air density ≅ 1.225 kg/m³
A is the frontal area of the car
[tex]C_{D}[/tex] is drag coefficient of the car
v is the relative velocity of air and the car, and will be taken as the initial velocity of the car before starting to decelerate.
imputing these values, we'll have
[tex]107.1 = \frac{1}{2}*1.225*2.35*C_{D}*20^{2}[/tex] = 575.75[tex]C_{D}[/tex]
[tex]C_{D}[/tex] = 107.1/575.75 = 0.189
