brycenrg said:
What takes more energy a car going 0 to 60 in 3 seconds or a car going 0 to 60 in 8 seconds.
brycenrg said:
Just talking about real life. I made up the problem
Since you just made up the problem to try to understand it better, I'll show you how to use equations to figure it out.
The equations you use are the kinematic equations for constant acceleration (assuming the cars have constant acceleration, and start from rest):
Velocity is acceleration multiplied by time: v(t) = a * t
Distance is related to the acceleration and time squared: d(t) = 1/2 a * t^2
Force is mass multiplied by acceleration: F = ma
And the Work done is the product of the force and the distance: W = F * d
Since both cars reach the final speed of 60m/h, we can use the velocity equation to find the ratio of their accelerations:
v(final) = 60 = a1 * t1 = a2 * t2
So a1/a2 = t2/t1 = 8/3
And we can take the ratio of the distances: [d1/d2] = [(1/2 a1` * t1^2) / (1/2 a2 * t2^2)] = [(a1 * 3^2) / (a2 * 8^2)] = (a1/a2) * (3^2/8^2) = (8/3) * (3/8)^2 = 3/8
Finally we can get the ratio of the forces F1/F2 = (m1 * a1)/(m2 * a2) = a1/a2 = 8/3 since the cars have equal mass.
So we can now calculate the ratio of the work done for the two accelerating cars:
W1/W2 = (F1 * d1) / (F2 * d2) = F1/F2 * d1/d2 = a1/a2 * d1/d2 = 8/3 * 3/8 = 1
So the quicker car accelerates over a shorter distance, but uses the same amount of energy as the car accelerating to the same speed over a longer distance (neglecting air resistance, etc.). Does that start to make some sense? And that's why the final Kinetic Energy KE for each is the same, since they have the same mass and the same velocity at the end.
