A subway car leaves station A; it gains speed at the rate of 4 ft/s^2 for of until it has reached and then at the rate 6 then th
e speed of 48 ft/s The car maintains the same speed until it approaches station B; driver applies the brakes, giving the car a constant deceleration and bringing it to a stop in 6 s. The total running time from A to B is 40 s Draw the a-t, v t, and s-t curves, and determine the distance between stations A and B.
1 answer:
Answer:
Refer to the attached document
1512 ft
Explanation:
Because the acceleration is constant or zero, the acceleration-time graph consists of horizontal segments. The values for t2 and a4 are derived as follows:
Acceleration - Time
0 < t < 6: Velocity change = area beneath the a–t graph
V_6 - 0 = (6 s)(4 ft/s²) = 24 ft/s
6 < t < t2: The velocity rises from 24 to 48 ft/s,
Velocity change = area beneath the a–t graph
48 - 24 = (t2 - 6) * 6
t2 = 10 s
t2 < t < 34: The velocity remains constant, meaning acceleration is zero.
34 < t < 40: Velocity change = area beneath the a–t graph
0 - 42 = 6*a4
a4 = - 8 ft / s²
A negative acceleration shows the area lies below the t-axis, indicating a decrease in speed.
Velocity - Time
Since acceleration remains constant or zero, the v−t graph is made up of linear segments connecting the calculated points.
Position change = area beneath the v−t graph
0 < t < 6: x6 - 0 = 0.5*6*24 = 72 ft
6 < t < 10: x10 - x6 = 0.5*4*(24 + 48) = 144 ft
10 < t < 34: x34 - x10 = 48*24 = 1152 ft
34 < t < 40: x40 - x34 = 0.5*6*48 = 144 ft
Summing these position changes yields the distance from A to B:
d = x40 - 0 = 1512 ft
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