Question

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.

Answer 1

a)

Step-by-step 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:

`x_(fc) = v_(o)*t + (1)/(2)*a*t^(2) (1)`

• 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:

`a_(c) =(v_(fc))/(t) = (20m/s)/(5s) = 4 m/s2 (2)`

• Replacing a and t in (1):

`x_(fc) = v_(o)*t + (1)/(2)*a*t^(2) = (1)/(2)*a*t^(2) = (1)/(2)* 4 m/s2*(5s)^(2) = 50.0 m. (3)`

• 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:

`a_(t) =(-v_(to))/(t) = (-20m/s)/(10s) = -2 m/s2 (4)`

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

`x_(ft) = 20m/s*10.0s + (1)/(2)*(-2 m/s2)*(10.0s)^(2) = 200m -100m = 100.0m (5)`

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