The vehicle motor or engine can be size based on required peak power. The energy or average power required provides a sense for how much fuel is required.

1. Calculate the average and peak power (kW) needed to accelerate a 1364 kg vehicle from 0 to 60 mph in 6 seconds. Assume that aerodynamic, rolling, and hill‐climbing force counts for an extra 10% of the needed acceleration force.
2. Chart the average and peak power (kW) vs. time duration from 2 to 15 seconds.
3. How much energy (kWh) is required to accelerate the vehicle?

1. Parg = 89.954 kw

  Pmax = 179.908 kw

2. Parg = 29.984 kw

  Pmax = 59.96 kw

3. Energy = 0.15 kWh

Explanation:

See the attached file for the calculation

1) P = 81.74 kW

2) As seen on the pic.

3) E = 0.1362 kWh

Explanation:

1) Given

m = 1364 kg (mass of the vehicle)

vi = 0 mph = 0 m/s  (initial speed)

vf = 60 mph = (60 mph)(1609 m/ 1 mi)(1 h/ 3600 s) = 26.8167 m/s  (final speed)

t = 6 s

We get the acceleration as follows

a = (vf - vi)/t  ⇒   a = (26.8167 m/s - 0 m/s)/ 6s

⇒   a = 4.469 m/s²

then the Force is

F = m*a ⇒   F = 1364 kg*4.469 m/s²

⇒   F = 6096.32 N

Then we get the average power as follows

P = F*v(avg) = F*(vi + vf)/2

⇒   P = 6096.32 N*(0 m/s + 26.8167 m/s)/2

⇒   P = 81741.59 W = 81.74 kW

Knowing that that aerodynamic, rolling, and hill‐climbing force counts for an extra 10% of the needed acceleration force, we use the following formula to find the peak power

Pmax = (1 + 0.1)*F*vf  = 1.1*F*vf

⇒   Pmax = 1.1*6096.32 N*26.8167 m/s

⇒   Pmax = 179831.503 W = 179.83 kW

2) The pic 1 shows the average and peak power (kW) vs. time duration from 2 to 15 seconds.

In the first chart we use the equation

Pinst = F*v  where F is constant and v is the instantaneous speed, and Pavg is the mean of the values.

In the second chart, we use the equation

Ppeak = 1.1*F*v  where F is constant and v is the instantaneous speed.

3) For 0 s ≤ t ≤ 6 s

We can use the equation

E = ΔK = Kf - Ki = 0.5*m*(vf² - vi²)

⇒   E = 0.5*1364 kg*((26.8167 m/s)² - (0 m/s)²)

⇒   E = 490449.123 J = (490449.123 J)(1 kWh/3.6*10⁶J)

⇒   E = 0.1362 kWh

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