Question

if x = t^3/3-5/2t^2 +6t +1 , then the value of d^2x / dt^2 , when dx/dt is zero

Answer 1

here is ur answer.......

Answer 2

Answer:

Step-by-step explanation:

The given equation is:

$x=\frac{t^{3}}{3}-\frac{5}{2}t^{2}+6t+1$

Now, differentiating the above equation wrt t, we get

$\frac{dx}{dt}=t^{2}-5t+6$

Also, $\frac{dx}{dt}=0$

⇒$t^{2}-5t+6=0$

⇒$t^{2}-2t-3t+6=0$

⇒$t(t-2)-3(t-2)=0$

⇒$(t-2)(t-3)=0$

⇒$t=2,3$

Now, again differentiating the equation wrt t, we get

$\frac{d^{2}x}{dt^{2}}=2t-5$

At t=2, $\frac{d^{2}x}{dt^{2}}=2(2)-5=-1$

At t=3, $\frac{d^{2}x}{dt^{2}}=2(3)-5=1$