Question

How to solve 2.14 using calculus

Answer 1

Answer:

The acceleration is $6.42\frac{m}{s^2}$

Explanation:

Given the velocity function:

$v=0.86\frac{m}{s^3}t^2$

you can obtain the instantaneous acceleration "a" as its first derivative:

$a=\dot{v}=2\cdot0.86\frac{m}{s^3}\cdot t=1.72\frac{m}{s^3}t$

To determine the value of "a" when the velocity was 12m/s, you need to figure out the value for "t" when this happens. At what time t is the velocity 12m/s?

$12.0\frac{m}{s}=0.86\frac{m}{s^3}t^2\implies t=3.74s$

This value of t is less than the 5 seconds mentioned in the text - so that is a good sign that the formula is valid for this value. And so you can use t=3.47s in the derivative (acceleration) above:

$a=1.72t=1.72\frac{m}{s^3}\cdot 3.74s = 6.42\frac{m}{s^2}$