Calculating Average Power for Accelerating Car

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To calculate the average power required to accelerate a 1200 kg car from rest to 20 m/s in 10 seconds, the kinetic energy formula KE = 1/2 mv^2 is applied, resulting in 240,000 joules of energy gained. Since power is defined as energy divided by time, the average power can be calculated by dividing the total energy by the time taken, yielding 24,000 watts. The initial kinetic energy is zero as the car starts from rest, and the time factor is crucial for determining average power. The calculations confirm that the average power output during the acceleration is indeed 24,000 watts. This approach effectively illustrates the relationship between energy, time, and power in physics.
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Hello Everyone. I am almost through my on-line coarse on Introductory Physics. I am taking my last self-assessment before the on-line certification in two weeks. There is just one problem that has my confused. Please provide some guidance. Thanks.

Accelerating Car - What is the average power required to accelerate a 1200 kg car from rest to a velocity of 20 m/s in 10 seconds?

I am not even sure exactly where to start. I know that power can be measured in joules/sec a metric unit known as a watt. I do know some related formulas that might apply. They include:

Power = Change in Energy/Change in time
F = ma
W = Fd

I am still unclear how to proceed to determine that answer to this question in watts. :eek:
 
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You have another:
KE=1/2mv^2.
 
How appropriate. If anyone can help me it is the master of falling (accelerating) bodies...Galileo. Anyway, I am still confused on why your formula is the one I should start with. Can you explain? If I do use this formula I get:

KE=(1/2)(1200kg)(20m/s)^2
KE=(.5)(1200)(400)
KE=240,000

Now, is this watts? Also, is it important to consider the 10 second time in calculating the avergae power? If so then maybe the answer should be 240,000/10=24,000 watts. What do you think? I am on tract? Thanks.
 
Looks good to me :approve:

The car is started at rest, so it had no kinetic energy.
After 10 seconds it had 240,000 J of kinetic energy.

So it gained 24.000 J per second on average.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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