Answer:
Part a)
[tex]v = 17.65 m/s[/tex]
Part b)
[tex]t = 25.22 s[/tex]
Part c)
[tex]d = 445.25 m[/tex]
Explanation:
Part a)
Speed of the truck is given as
[tex]v = 39.5 mph[/tex]
also we know that
1 mile = 1609 m
1 h = 3600 s
now we have
[tex]v = 39.5 \times \frac{1609}{3600}[/tex]
[tex]v = 17.65 m/s[/tex]
Part b)
Now if car catches the truck after time "t"
so the distance moved by car and truck will be same in this time
[tex]\frac{1}{2}at^2 = v t[/tex]
[tex]\frac{1}{2}1.4 t^2 = 17.65 t[/tex]
[tex]t = 25.22 s[/tex]
Part c)
Distance moved by the car in the same time
[tex]d = \frac{1}{2}at^2[/tex]
[tex]d = \frac{1}{2}(1.4)(25.22^2)[/tex]
[tex]d = 445.25 m[/tex]