On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a constant rate from 20 m/s to a complete stop over a 10 second interval. How does the distance traveled by the truck compare to that of the car? So assuming that I've sketched a graph, do I just do distance x time? a. The truck travels twice as far as the car.
b. There is not enough information to answer the question.
c .The truck travels the same distance as the car.
d The truck travels half as far as the car.

a)

Explanation:

Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

       

Since the car starts from rest, v₀ =0.We know the value of t = 5 sec., but we need to find the value of a.Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       

Replacing a and t in (1):

       

Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       

Replacing v₀, at and t in (1), we have:

       

Therefore, as the truck travels twice as far as the car, the right answer is a).

Kiloleter, mega leterhope this !

"induction" is the process of charging an object without touching it.hope my answer has you : )