Answer:
The acceleration is [tex]6.42\frac{m}{s^2}[/tex]
Explanation:
Given the velocity function:
[tex]v=0.86\frac{m}{s^3}t^2[/tex]
you can obtain the instantaneous acceleration "a" as its first derivative:
[tex]a=\dot{v}=2\cdot0.86\frac{m}{s^3}\cdot t=1.72\frac{m}{s^3}t[/tex]
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?
[tex]12.0\frac{m}{s}=0.86\frac{m}{s^3}t^2\implies t=3.74s[/tex]
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:
[tex]a=1.72t=1.72\frac{m}{s^3}\cdot 3.74s = 6.42\frac{m}{s^2}[/tex]
