Question

An object moves horizontally along a line. Its position from a fixed point can be modelled by s(t) = t^3 - 6t^2 - 15t + 20, where s is the displacement, in cm, and t is the time. in seconds. Determine the object's maximum velocity on the interval 0 ≤ t ≤ 5.

(I got an answer of 0 cm per second as the maximum velocity, just checking if anyone else has got the same thing) (Calculus - Optimisation)

Answer 1

ok so u got\[v(t)=\frac{ds}{dt}=3t^2-12t-15\]and u want to calculate the maximum velocity , actually u must find the maximum of \(|v(t)|\)

Answer 2

what is the maximum of |v| on the interval 0 ≤ t ≤ 5

Answer 3

I first took the derivative of the displacement equation to get v(t) =3t^2−12t−15 and factorised it to get t(t-5)(t-5) = 0, but we still get 0 and 5, which are part of the domain of t (0 ≤ t ≤ 5). So I then took the second derivative of s(t) and got s(t)'' = 6t - 12.

Answer 4

And if you simplify that, you get t = 2

Answer 5

and if you put all those values into the first derivative, you get -15cm/s for t = 0
-27cm/s for t = 2
and 0cm/s for t = 5

Answer 6

good :) and so what is the maximum velocity?

Answer 7

That is where I am confused, is it -27cm/s because it has turned around and is moving left or is it just 0cm/s if it moves to the right?

Answer 8

u got the point.. answer is -27 cm/s... here the negative sign shows the direction of the object

Answer 9

Okay, thank you :)

Answer 10

and just note that\[3t^2-12t-15=3(t^2-4t-5)=3(t+1)(t-5)\]answer is still -27 cm/s
good luck :)