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2 years ago

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A Porsche 944 Turbo has a rated engine power of 217hp . 30% of the power is lost in the drive train, and 70% reaches the wheels.

The total mass of the car and driver is 1520 kg, and two-thirds of the weight is over the drive wheels.
a) What is the maximum acceleration of the Porsche on a concrete surface where μs=1?

b) What is the speed of the Porsche at maximum power output?

c) If the Porsche accelerates at amax , how long does it take until it reaches the maximum power output?

1 answer:

scZoUnD [109]2 years ago

6 0

Explanation:

(a) It is given that two-third of weight is over the drive wheels. So, mathematically, w = $\frac{2}{3}mg$ .

Hence, maximum force is expressed as follows.

$F_{max} = \mu_{s} \times w$

$m \times a_{max} = \mu_{s} (\frac{2}{3} mg)$

Hence, the maximum acceleration is calculated as follows.

$a_{max} = \frac{2}{3} \mu_{s} \times g$

= $\frac{2}{3} \times 1.00 \times 9.8 m/s^{2}$

= 6.53 $m/s^{2}$

Hence, the maximum acceleration of the Porsche on a concrete surface where μs = 1 is 6.53 $m/s^{2}$ .

(b) Since, 30% of the power is lost in the drive train. So, the new power is 70% of $P_{max}$ .

That is, new power = $0.7 \times P_{max}$

Now, the expression for power in terms of force and velocity is as follows.

P = $F_{max} \nu$

$0.7 P_{max} = ma_{max} \nu$

Therefore, speed of the Porsche at maximum power output is as follows.

$\nu = 0.7 \times \frac{P_{max}}{ma_{max}}$

= $0.7 \times \frac{217 hp \times \frac{746 W}{1 hp}}{1500 kg \times 6.53 m/s^{2}}$

= 11.568 m/s

= 11.57 m/s

Therefore, speed of the Porsche at maximum power output is 11.57 m/s.

(c) The time taken will be calculated as follows.

time = $\frac{\text{velocity}}{\text{acceleration}}$

= $\frac{11.57 m/s}{6.53 m/s^{2}}$

= 1.77 s

Therefore, the Porsche takes 1.77 sec until it reaches the maximum power output.

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