Answer:
1) P = 81.74 kW
2) As seen on the pic.
3) E = 0.1362 kWh
Step-by-step 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