Question

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 the 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.

Answer 1

Answer:

See attachment

1512 ft

Explanation:

Since the acceleration is either constant or zero,  the a−t curve is made of horizontal straight-line segments. The values of t2  and a4 are determined as follows:

Acceleration - Time

0 < t < 6: Change in v = area under a–t curve

V_6 - 0 =  (6 s)(4 ft/s2 ) = 24 ft/s

6 < t < t2: Since the velocity increases from 24 to 48 ft/s,

Change in v = area under a–t curve

48 - 24 = (t2 - 6) * 6

t2 = 10 s

t2 < t < 34: Since the velocity is constant, the acceleration is zero.

34 < t < 40: Change in v = area under a–t curve

0 - 42 = 6*a4

a4 = - 8 ft / s^2

The acceleration being negative, the corresponding area is below the t axis;  this area represents a decrease in velocity.

Velocity - Time

Since the acceleration is either constant or zero, the  v−t curve is made of straight-line segments connecting the points determined above.

Change in x = area under v−t curve

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

Adding the changes in x, we obtain the distance from A to B:

d = x40 - 0 = 1512 ft