Question

At time t ≥ 0, the velocity of a body moving along the s-axis is v = t² -5t +4. When is the body moving backwards

Answer 1

Answer:

A

Step-by-step explanation:

The velocity of a moving body is given by the equation:

$v=t^2-5t+4 ,\, t\geq0$

Is the velocity is positive (v>0), then our object will be moving forwards.

And if the velocity is negative (v<0), then our object will be moving backwards.

We want to find between which interval(s) is the object moving backwards. Hence, the second condition. Therefore:

$v$

By substitution:

$t^2-5t+4$

Solve. To do so, we can first solve for t and then test values. By factoring:

$(t-4)(t-1)=0$

Zero Product Property:

$t=1, \text{ and } t=4$

Now, by testing values for t<1, 1<t<4, and t>4, we see that:

$v(0)=4>0,\, v(2)=-20$

So, the (only) interval for which v is <0 is the second interval: 1<t<4.

Hence, our answer is A.