Question

A car rolls down a ramp in a parking garage. The horizontal position of the car in meters over time is shown below. Graph of vertical position (in meters) on y axis and time (in seconds) on x axis. The initial position is 12 m at t=0 s, and the position decreases linearly to 6 m at t=8 s. Then the position is constant at 6 m until t=16 s, then decreases linearly to 0 m at t=24 s. Graph of vertical position (in meters) on y axis and time (in seconds) on x axis. The initial position is 12 m at t=0 s, and the position decreases linearly to 6 m at t=8 s. Then the position is constant at 6 m until t=16 s, then decreases linearly to 0 m at t=24 s. What is the displacement of the car between 0\text{ s}0 s0, start text, space, s, end text and 24\text{ s}24 s24, start text, space, s, end text? 6 \text mmstart text, m, end text What is the distance traveled by the car between 0\text{ s}0 s0, start text, space, s, end text and 24\text{ s}24 s24, start text, space, s, end text?

Answer 1

Answer:

 d_total = 12 m

Explanation:

This is a kinematics problem, in the attachment we can see the problem graph, they ask us how much the distance is worth for all the time of 24 s.

The total distance is

            d_total = d₁ + d₂ + d₃

as in the graph the part of d₂ is horizontal (v=0) the distance is zero d₂ = 0

the distance d₁ is

         d₁ = 12 -6 = 6 m

the distance d₃

         d₃ = 6 -0 = 6 m

therefore the total distance traveled is

         d_total = 6 + 0 + 6

         d_total = 12 m