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

Answer:

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:

        $x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} (1)$

• 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:

       $a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2 (2)$

• Replacing a and t in (1):

       $x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m. (3)$

• 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:

       $a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2 (4)$

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

       $x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m (5)$

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