Question

A car moves along an x axis through a distance of 950 m, starting at rest (at x = 0) and ending at rest (at x = 950m). Through the first 1/4 of that distance, its acceleration is +5.40 m/s^2. Through the next 3/4 of that distance, its acceleration is -1.80 m/s^2. What are (a) its travel time through the 950 m and (b) its maximum speed?

Answer 1

Explanation:

To find the time it takes for the car to travel 950 m, we need to find the time it takes for each segment and add them up. Let's call the travel time for the first quarter t1, and the travel time for the remaining 3 quarters t2.

a) Travel time:

For the first quarter (0 m to 237.5 m), the equation for the velocity can be found using the kinematic equation:

v = v0 + at

where v0 is the initial velocity (0 m/s), a is the acceleration (+5.40 m/s^2), and t is t1.

So,

t1 = (v - v0) / a = (0 - 0) / 5.40 = 0

For the next 3 quarters (237.5 m to 950 m), the equation for the velocity can be found using the kinematic equation:

v^2 = v0^2 + 2ad

where d is the displacement (950 m - 237.5 m = 712.5 m), and a is the acceleration (-1.80 m/s^2).

So,

t2 = sqrt((2ad + v0^2) / a) = sqrt((2 * -1.80 * 712.5 + 0^2) / -1.80) = sqrt(1286.25 / -1.80) = sqrt(709.03) = 26.6 s

Adding t1 and t2 gives the total travel time:

t = t1 + t2 = 0 + 26.6 = 26.6 s

b) Maximum speed:

To find the maximum speed, we need to find the velocity at the halfway point (475 m) by using the kinematic equation:

v = v0 + at

where v0 is the initial velocity (0 m/s), a is the acceleration (+5.40 m/s^2), and t is t1.

So,

v = 0 + 5.40 * (475 / 237.5) = 4.96 m/s

The maximum speed is 4.96 m/s.

sorry for making it so long.