Two students are discussing how the speed of the car compares to the speed of the truck when both vehicles are in front of the house. Student 1 says, "The distance traveled by the car and the truck is the same, and the time is the same, so they must have the same speed." Student 2 says, "I don’t see how that can be. The car catches up to the truck, so the car has to be going faster."

a. Which aspects of Student 1’s reasoning, if any, are correct? Support your answer in terms of relevant features of your graphs in part (a).
b. Which aspects of Student 2’s reasoning, if any, are correct? Support your answer in terms of relevant features of your graphs in part (a).
c. Derive an expression for the acceleration of the car. Express your answer in terms of D and vt
d. Determine the time at which the speed of the car is equal to the speed vt of the truck. Express your answer in terms of tD. Justify your answer.

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moya1316 moya1316 · Answer 1 · 2020-10-13T11:23:53+02:00

Answer:

a) Student 1 is right in the way he calculates the speed cid d of the truck, since going at constant speed

b) Student 2 is right that for the car to reach the truck it must have some relationship,

c) t = √d/a

Explanation:

In this exercise you are asked to analyze the movement in a mention using kinematics.

a) Student 1 is right in the way he calculates the speed cid d of the truck, since going at constant speed he can use the equation

v = d / t

b) Student 2 is right that for the car to reach the truck it must have some relationship, which is given by

v = v₀ + a t

x = v₀ t + ½ a t²

c) let's use the equations

v = d / t

v = v₀ + at

d / t = vo + at

If we can assume that the car starts from rest, so v₀ = 0

d / t = a t

t² = d / a

t = √d / a

if the speed of the car is not zero the result is a little more complicated

d = vo t + a t²

at² + vo t - d = 0

let's solve the quadratic equation

t = [-v₀ ± √ (v₀² + 4a d)] / 2a