A particle, P, moves along the x-axis. At time t seconds, t > 0, the displacement, is given by x=1/2t^2(t ^2−2t+1).

Find the times when is instantaneously at rest.In order to solve this question we first have to multiply out in order to obtain the full expression of x which will be x = 1/2t^4 -2t^3+1/2t^2. Now we differentiate with respect to time we obtain v=2t^3 -3t^2+t. If P is suppose to be at rest then v will be equal zero. So we obtain an equation 0=2t^3-3t^2+t and solving the equation t(2t-1)(t-1)=0 and we obtain three different answers t=0, t=1/2 and t=1 and all answers are possible.

AK

Related Maths A Level answers

All answers ▸

Use the substitution u = 6 - x^2 to find the value of the integral of (x^3)/(sqrt(6-x^2)) between the limits of x = 1 and x = 2 (AQA core 3 maths


If (m+8)(x^2)+m=7-8x has two real roots show that (m+9)(m-8)<0 where m is an arbitrary constant


Find the general solution of 2 dy/dx - 5y = 10x


How can I find the area under the graph of y = f(x) between x = a and x = b?