Physics: A car accelerates uniformly from rest to 24.8 m/s in 7.88 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 8.55 x 103 N, and (b) the weight of the car is 1.10 x 104 N.

A car accelerates uniformly from rest to 24.8 m/s

elwinelwin9475

22.01.2022

When he weight of the car is 8.55 x $10^{3}$ N then power = 314.012 KW

When he weight of the car is 1.10 x $10^{4}$ N then power = 43.76 KW

Explanation:

Given that

Initial velocity $V_{1}$ = 0

Final velocity $V_{2}$ = 24.8 $\frac{m}{s}$

Time = 7.88 sec

We know that power required to accelerate the car is given by

P = $\frac{change \ in \ kinetic \ energy}{time}$

Change in kinetic energy Δ K.E = $\frac{1}{2} m (V_{2}^{2} - V_{1}^{2} )$

Since Initial velocity $V_{1}$ = 0

⇒ Δ K.E = $\frac{1}{2} m V_{2} ^{2}$

⇒ Power P = $\frac{1}{2} \frac{m}{t} V_{2} ^{2}$

⇒ Power P = $\frac{1}{2} \frac{W}{g\ t} V_{2} ^{2}$ -------- (1)

(a). The weight of the car is 8.55 x $10^{3}$ N = 8550 N

Put all the values in above formula

So power P = $\frac{1}{2} \frac{8550}{(9.81)\ (7.88)} (24.8) ^{2}$

P = 314.012 KW

(b). The weight of the car is 1.10 x $10^{4}$ N = 11000 N

Put all the values in equation (1) we get

P = $\frac{1}{2} \frac{11000}{(9.81)\ (7.88)} (24.8) ^{2}$

P = 43.76 KW

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