Question

Question 30 (10 points) A bicycle rider rides away from home along a highway and back along the same road in such a way that her distance from home at time t is given by x(t) t4 8(Equation:eqn t3 ,heights 32,width-24.size-14.bgcolor: #FFFFFF.title-eqn_2.controls-false}+ 16(Equation:eqn t2 ,height-32width-24.size-14.bgcolor: #FFFFFF.title-eqn_3.controls-false) where t is in hours and x is in kilometers When does her maximum speed occur? 22/(3) hours out 2 hours out 2 ± 1/2 hours out 1 and 3 hours out

Answer 1

Given that her distance with time is-

$\mathbf{x(t)=t^{4}-8t^{3}+16t^{2}}$

where t is in hours and x is in kilometers.

To find the time at which maximum speed occurs,we will determine the velocity function with time.

$v(t)=\frac{d}{dt}x(t)=\frac{d(t^{4}-8t^{3}+16t^{2})}{dt}$

$\Rightarrow v(t)=4t^{3}-24t^{2}+32t ------------(1)$

now to find the time of maximum velocity we will use the principle of maxima-minima by using calculus.

Differentiate equation(1) with respect to t we get-

$\Rightarrow \frac{d(v(t))}{dt}=12t^{2}-48t+32 ------------(2)$

Now equating equation(2) with zero and find the value of t-

$\Rightarrow \frac{d(v(t))}{dt}=12t^{2}-48t+32=0$

$\Rightarrow 3t^{2}-12t+8=0$

here a=3,b=-12 and c=8

using quadratic equation formula to find the values of t.

$t=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$\Rightarrow t=\frac{12\pm \sqrt{144-96}}{6}$

$\Rightarrow t=\frac{12\pm 4\sqrt{3}}{6}$

$\Rightarrow t=2\pm \frac{ 2\sqrt{3}}{3}$  

so first option is correct.