If a 1300-kg car can accelerate from 35 kmh to 65 kmh in 3.8 s, how long will it take to accelerate from 55 kmh to 95 kmh? Assume the power stays the same, and neglect frictional losses.

6.76 s

Explanation:

Power = Energy/time

P = E/t Equation 1

Where P = Power, E = kinetic energy, t = time.

But,

E = 1/2m(Δv)² Equation 2

Where m = mass of the car, Δv = change in velocity.

Substitute equation 2 into equation 1

P = 1/2m(Δv)²/t Equation 3

Given: m = 1300 kg, Δv = 65-35 = 30 km/h = 30(1000/3600) m/s = 8.33 m/s, t = 3.8 s

Substitute into equation 3

P = 1/2(1300)(8.33²)/3.8

P = 11869.15 W.

Assuming the power stays the same,

making t the subject of formula in equation 3

t = 1/2m(Δv)²/P Equation 4

Given: m = 1300 kg, Δv = 95-55 = 40 km/h = 11.11 m/s, P = 11869.15 W

Substitute into equation 4

t = 1/2(1300)(11.11²)/11869.15

t = 6.76 s

explanation: in any substance/object, the particles inside it (atoms/molecules) constantly move in random directions and with random speeds (this motion is called brownian motion). as a result, the particles have some kinetic energy (which is proportional to the temperature of the substance).

explanation:

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scientific laws are the result of testing theories and proving them to be facts.