Question

Tesla Model S and the driver have a total mass of 2250 kg. The projected front area of the car is 2.35 m2. The car is traveling at 72km/hr when the driver puts the transmission into neutral and allows the car to freely coast until after 105 seconds its speed reaches 54 km/hr. Determine the drag coefficient for the car, assuming its values is constant. Neglect rolling and other mechanical resistance.

Answer 1

The drag coefficient of the car is 0.189

Step-by-step 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

`a = (v-u)/(t)`

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

`a = (15-20)/(105)` = -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..

`F_(D) = ma`

where
`F_(D)` is the drag force

m is the mass of the car

a is the deceleration of the car

imputing values, we'll have

`F_(D) = 2250*0.0476` = 107.1 N

equation for drag force is

`F_(D) = (1)/(2)pAC_(D) v^(2)`

where p is the air density ≅ 1.225 kg/m³

A is the frontal area of the car

`C_(D)` 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

`107.1 = (1)/(2)*1.225*2.35*C_(D)*20^(2)` = 575.75
`C_(D)`

`C_(D)` = 107.1/575.75 = 0.189