Starting at t = 0 , the acceleration of a cart varies with time as shown in the figure. Determine the distance travelled in 30 seconds and select the correct description of its position-time graph.

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Quantrex Academy · Accepted Answer

Correct answer: B. 1000 ft ; the position-time graph is parabolic for 0 to 10 s and 20 to 30 s , and a sloped straight line for 10 to 20 s .

Assuming the cart starts from rest, the initial velocity is

u = 0

.

For the time interval

0 \le t \le 10 \text{ s}

:
Acceleration

a_1 = 5 \text{ ft/s}^2

.
Velocity at

t = 10 \text{ s}

is

v_1 = u + a_1 t = 0 + 5 \times 10 = 50 \text{ ft/s}

.
Distance travelled

s_1 = \dfrac{1}{2} a_1 t^2 = \dfrac{1}{2} \times 5 \times (10)^2 = 250 \text{ ft}

.
Since acceleration is constant and non-zero, the position-time graph is a parabola.

For the time interval

10 < t \le 20 \text{ s}

:
Acceleration

a_2 = 0 \text{ ft/s}^2

.
The cart moves with a constant velocity

v_2 = 50 \text{ ft/s}

.
Distance travelled

s_2 = v_2 \times \Delta t = 50 \times 10 = 500 \text{ ft}

.
Since velocity is constant, the position-time graph is a sloped straight line.

For the time interval

20 < t \le 30 \text{ s}

:
Acceleration

a_3 = -5 \text{ ft/s}^2

.
Distance travelled

s_3 = v_2 \Delta t + \dfrac{1}{2} a_3 (\Delta t)^2 = 50 \times 10 - \dfrac{1}{2} \times 5 \times (10)^2 = 500 - 250 = 250 \text{ ft}

.
Since acceleration is constant and non-zero, the position-time graph is again a parabola.

Total distance travelled

S = s_1 + s_2 + s_3 = 250 + 500 + 250 = 1000 \text{ ft}

.

Starting at t = 0 , the acceleration of a cart varies with time as...

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