Date: 07/14/99 at 13:39:03 From: Kirsten Peterson Subject: Moving Particle Equation In my College Calculus class we were given the following problem. I think I understand the logic behind it but the answers just don't seem right. Could you please help me? This is the problem: Assume a particle moves on the x-axis according to the formula x = t^3-6t^2+9t+5 Find: a. the velocity when t = 3 b. the acceleration when t = 4 c. the times when the velocity is zero d. the place where the acceleration is zero e. the acceleration at each of the two moments where the particle is motionless This is what I did: First I differentiated the equation dx/dt = 3t^2-12t+9 a. velocity when t = 3 I then plugged in 3 for t dx/dt = 3(3)^2-12(3)+9 = 0 b. I do not remember how to find the acceleration of a particle. c. logically, with how I did a, the answer to this problem would be the given in a, because the answer for the velocity is 0 so the answer to this equation is going to be t = 3. This is where my logic doesn't make any sense. How can that be the right answer? I pretty much stopped there because I got confused. Please help me? Thank you! Kirsten
Date: 07/14/99 at 16:32:56 From: Doctor Anthony Subject: Re: Moving Particle Equation >Assume a particle moves on the x-axis according to the formula > x=t^3-6t^2+9t+5 >Find: >a. the velocity when t=3 dx/dt = 3t^2 - 12t + 9 and putting t = 3 dx/dt = 27 - 36 + 9 = 0 so the particle is stationary at t = 3. >b. the acceleration when t = 4 accel = d^2(x)/dt^2 = 6t - 12 and putting t = 4 accel = 24 - 12 = 12 m/sec^2 >c. the times when the velocity is zero. We solve 3t^2 - 12t + 9 = 0 t^2 - 4t + 3 = 0 (t-1)(t-3) = 0 So the particle is stationary at t = 1 and t = 3. >d. the place where the acceleration is zero. 6t - 12 = 0 so accel = 0 when t = 2 and then x = t^3 - 6t^2 + 9t + 5 = 8 - 24 + 18 + 5 = 7 metres >e. the acceleration at each of the two moments where the particle is >motionless. When t = 1 accel = 6t - 12 = -6 m/sec^2 when t = 3 accel = 16 - 12 = +6 m/sec^2 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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