# In which case the car engine must do a lot of work to accelerate from standstill to a speed of 27 km / h

In which case the car engine must do a lot of work to accelerate from standstill to a speed of 27 km / h or to increase the speed from 27 to 54 km / h.

V0 = 0 m / s.
V1 = 27 km / h = 7.5 m / s.
V2 = 54 km / h = 15 m / s.
A -?
Car engine operation. A is determined by the formula: A = F * S, where F is the engine thrust force, S is the engine displacement.
Let us express the thrust force of the engine according to 2 Newton’s laws: F = m * a, where m is the mass of the car, a is the acceleration of the car.
Let’s write the formula for acceleration: a = (V – V0) / t.
A = m * (V – V0) * S / t = m * ΔV * S / t.
Since the change in speed is the same: ΔV1 = V1 – V0 = 7.5 m / s – 0m / s = 7.5 m / s, ΔV2 = V21 – V1 = 15 m / s – 7.5 m / s = 7, 5 m / s, then work A will be determined by the distance S and the acceleration time t.
The greater the S / t ratio, the more work A.
Answer: the greater the ratio of the acceleration distance S to the acceleration time t, the more work A.

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