A 1400-kg race car can go from 0 to 90km/h in 6.0s . What average power is required to do this?
V = 90km/h = 90000m/3600s = 25 m/s.
a = (V-Vo)/t = (25-0)/6 = 4.167 m/s^2
d = 0.5a*t^2 = 0.5*4.167*6^2 = 75 m.
P = F*d/t = mg * d/t
P = 13720 * 75/6 = 171,500 J/s.
To find the average power required to accelerate the race car from 0 to 90 km/h in 6.0 seconds, we can use the formula:
Power = Work / Time
First, we need to calculate the work done. The work done to accelerate an object is given by the equation:
Work = (1/2) * m * v^2
where "m" is the mass of the object and "v" is the final velocity.
Given:
Mass of the race car (m) = 1400 kg
Final velocity (v) = 90 km/h = 25 m/s (1 km/h = 1/3.6 m/s)
Now, we can substitute these values into the formula:
Work = (1/2) * 1400 kg * (25 m/s)^2
= (1/2) * 1400 kg * 625 m^2/s^2
= 437500 J (Joules)
Next, we can calculate the average power by dividing the work by the time:
Power = Work / Time
= 437500 J / 6.0 s
= 72916.67 W (Watts)
Therefore, the average power required to accelerate the race car is approximately 72916.67 Watts.
To determine the average power required to accelerate a race car from 0 to 90 km/h, we can use the formula:
Power = Work / Time
In this case, the work done on the car is equal to the change in its kinetic energy:
Work = ΔKE = ½ * m * (vf^2 - vi^2)
Where:
m = mass of the car = 1400 kg
vf = final velocity = 90 km/h = 90 * (1000/3600) m/s
vi = initial velocity = 0 m/s
Plugging in the values into the formula, we have:
ΔKE = ½ * 1400 kg * ((90 * (1000/3600))^2 - 0^2)
Simplifying the expression:
ΔKE = ½ * 1400 kg * ((25 m/s)^2 - 0^2)
ΔKE = ½ * 1400 kg * (625 m^2/s^2)
ΔKE = ½ * 1400 kg * 625 m^2/s^2
ΔKE = 437500 J
Now, we need to calculate the average power using the given time:
Time = 6.0 s
Using the formula P = ΔKE / Time:
P = 437500 J / 6.0 s
P ≈ 72916.67 W
Therefore, the average power required to accelerate the race car from 0 to 90 km/h in 6.0 seconds is approximately 72,916.67 Watts.