Question

A body moves s metres in a time t seconds so that s = t3 – 3t2 + 8. Find:

Answer 1

Using derivatives, it is found that:

i) $v(t) = 3t^2 - 6t$

ii) 9 m/s.

iii) $a(t) = 6t - 6$

iv) 6 m/s².

v) 1 second.

What is the role of derivatives in the relation between acceleration, velocity and position?

• The velocity is the derivative of the position.

• The acceleration is the derivative of the velocity.

In this problem, the position is:

$s(t) = t^3 - 3t^2 + 8$

item i:

Velocity is the derivative of the position, hence:

$v(t) = 3t^2 - 6t$

Item ii:

$v(3) = 3(3)^2 - 6(3) = 27 - 18 = 9$

The speed is of 9 m/s.

Item iii:

Derivative of the velocity, hence:

$a(t) = 6t - 6$

Item iv:

$a(2) = 6(2) - 6 = 6$

The acceleration is of 6 m/s².

Item v:

t for which a(t) = 0, hence:

$6t - 6 = 0$

$6t = 6$

$t = \frac{6}{6}$

$t = 1$

Hence 1 second.

You can learn more about derivatives at brainly.com/question/14800626