Question

The acceleration of a cart started at t = 0, varies with time as shown in figure (3-E2). Find the distance travelled in 30 seconds and draw the position-time graph.

Answer 1

In the first 10 seconds,
${\mathrm{S}}_1=ut+\frac{1}{2}a{t}^2$$=0+\frac{1}{2}5\times {10}^2=250\mathrm{ft}$

At t = 10 s,
v = u + at = 0 + 5 × 10 = 50 ft/s
∴ From 10 to 20 seconds (∆t = 20 − 10 = 10 s), the particle moves with a uniform velocity of 50 ft/s.
Distance covered from t = 10 s to t = 20 s:
S₂ = 50 × 10 = 500 ft

Between 20 s to 30 s, acceleration is constant, i.e., −5 ft/s².
At 20 s, velocity is 50 ft/s.
t = 30 − 20 = 10 s
$$\begin{gathered} {\mathrm{S}}_3=ut+\frac{1}{2}a{t}^2=50\times 10+\frac{1}{2}\left(-5\right){10}^2 \\ \Rightarrow {\mathrm{S}}_3=500-250=250\mathrm{ft} \end{gathered}$$

Total distance travelled is 30 s:
S₁ + S₂ + S₃
= 250 + 500 + 250
= 1000 ft

The position–time graph: