Answer:
A
Step-by-step explanation:
The velocity of a moving body is given by the equation:
[tex]v=t^2-5t+4 ,\, t\geq0[/tex]
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:
[tex]v<0[/tex]
By substitution:
[tex]t^2-5t+4<0[/tex]
Solve. To do so, we can first solve for t and then test values. By factoring:
[tex](t-4)(t-1)=0[/tex]
Zero Product Property:
[tex]t=1, \text{ and } t=4[/tex]
Now, by testing values for t<1, 1<t<4, and t>4, we see that:
[tex]v(0)=4>0,\, v(2)=-2<0,\,\text{ and } v(5)=4>0[/tex]
So, the (only) interval for which v is <0 is the second interval: 1<t<4.
Hence, our answer is A.
