The position of the front bumper of a test car undermicroprocessor control is gi

Question

The position of the front bumper of a test car undermicroprocessor control is given by
x(t)=2017m +(4.80m/s^2)t^2-(0.100 m/s^6)t^6. (a) Find its position and acceleration at the instants whenthe car has zero velocity. *The solution to this problem on this cite is not veryclear. I don't understand how t^4= 4.80/0.600 So, please show me how to do it step by step with out skippinganything.! Thanks. I will rate high. The position of the front bumper of a test car undermicroprocessor control is given by
x(t)=2017m +(4.80m/s^2)t^2-(0.100 m/s^6)t^6. (a) Find its position and acceleration at the instants whenthe car has zero velocity. *The solution to this problem on this cite is not veryclear. I don't understand how t^4= 4.80/0.600 So, please show me how to do it step by step with out skippinganything.! Thanks. I will rate high.

Explanation / Answer

The velocity is the 1st derivative of the position function withrespect to time. so v(t)=x(t)'=2*4,8*t-6*0,1*t^5 the acceleration is the 2nd derivative of the velocityfunction, a(t)=v(t)'=2*4,8-6*5*0,1*t^4 when the velocity is zero, so v(t)=0=2*4,8t-6*0,1*t^5 4,8=3*0,1t^4 so t^4=4,8/0,3=16 so t=2(s) plug this in we have x and a x=2029,8 (m) a=-38,4(m/s2)

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