A motorist driving a 1400 kg car on level ground accelerates from 20.0 m/s to 30.0 m/s in a time of 6s.

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A motorist driving a 1400 kg car on level ground accelerates from 20.0 m/s to 30.0 m/s in a time of 6s.

[From: ] [author: ] [Date: 11-12-14] [Hit: ]

P = ½m/t (v² - u²) .. .. ½ x 1400 / 6 (30² - 20²) ........

Neglecting friction and air resistance, determine the average mechanical power in watts the engine must supply during this time interval.

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Power = rate of transfer of energy (into KE) = ∆E / t .. J/s (W)

P = (½mv² - ½mu²) / t
P = ½m/t (v² - u²) .. .. ½ x 1400 / 6 (30² - 20²) .. ►P(av) = 58 333.30 J/s .. (58.33 kW)

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