A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in watts required to accelerate the sports car from 30 m/s to 60 m/s in 9 seconds

First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:

[tex]W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]

where :

[tex]K_f = \frac{1}{2}mv^2[/tex] is the final kinetic energy of the car, with

m = 2000 kg is the mass of the car

v = 60 m/s is the final speed of the car

[tex]K_i = \frac{1}{2}mu^2[/tex] is the initial kinetic energy of the car, with

u = 30 m/s is initial speed of the car

Soolving:

[tex]W=\frac{1}{2}(2000)(60)^2 - \frac{1}{2}(2000)(30)^2=2.7\cdot 10^6 J[/tex]

Now we can find the power required for the acceleration, which is given by

[tex]P=\frac{W}{t}[/tex]

where

t = 9 s is the time elapsed

Solving:

[tex]P=\frac{2.7\cdot 10^6}{9}=3.0\cdot 10^6 W[/tex]

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codiepienagoya codiepienagoya · Answer 2 · 2022-02-22T13:31:41+01:00

The power is "300,000 W".

Power calculation:

[tex]\text{Mass} \ m= 2000\ kg\\\\ \text{intial velocity}\ u= 30 \frac{m}{s}\\\\ \text{final velocity}\ v= 60 \frac{m}{s}\\\\\text{Power}=?[/tex]

Using the formula for calculating the initial kinetic energy:

[tex]\to (K.E)_1 = \frac{1}{2}\ mu^2\\\\[/tex]

[tex]\to (K. E)_1 =\frac{1}{2}\times 2000\times 30^2[/tex]

[tex]= 900,000 \ J\\\\[/tex]

Using the formula for calculating the final kinetic energy:

[tex]\to (K.E)^2 = \frac{1}{2} m v^2 \\\\[/tex]

[tex]\to (K. E.)_2 = 1\times 2000 \times 60^2 \\\\[/tex]

[tex]=3600,000\ J\\\\[/tex]

Using the formula for calculating the power:

Power[tex]P= \frac{\text{Change in kinetic energy}}{time}\\\\[/tex]

[tex]\to \frac{(KE)_2-(k.E)_2}{time} =\frac{3600000 - 900000}{9}\\\\[/tex]

[tex]= \frac{2700000}{9} \\\\= 300,000\ W \\\\[/tex]

Therefore, the final answer is "300,000 W".

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