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Distance travelled by a car considering only air friction?
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[QUOTE="jack action, post: 6811865, member: 240508"] Here is how you arrive at the equation you used: The work done ##W## is defined as: $$W = \int Fdx$$ Knowing that ##F=ma##, the work done based on acceleration is: $$W= \int madx$$ If we want to know the work based on velocity alone: $$W = \int madx = \int m\frac{dv}{dt}dx = \int m\frac{dx}{dt}dv = \int mvdv$$ Both equations should give the same amount of work, so: $$\int_{v_i}^{v_f} mvdv = \int_{x_i}^{x_f} madx$$ $$\frac{1}{2}m(v_f^2 - v_i^2) = ma(x_f - x_i)$$ $$v_f^2 = v_i^2 + 2a(x_f - x_i)$$ Which is the equation you used. This assumes that ##F = ma##, where ##a## is constant. But that is not the case here. The problem identifies the force ##F## that you need to use with the work ##Fdx##. And the work done based on velocity is still ##mvdv##. All you need to do is to equate both as done previously and resolve the integrals. [[B]Hint:[/B] the air friction force varies with velocity.] [/QUOTE]
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Distance travelled by a car considering only air friction?
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