Short Note

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

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#### Solution

In the first 10 seconds,

\[S_1 = ut + \frac{1}{2}a t^2\]

\[= 0 + \frac{1}{2}5 \times {10}^2 = 250 ft\]

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:

S

Between 20 s to 30 s, acceleration is constant, i.e., −5 ft/s

At 20 s, velocity is 50 ft/s.

t = 30 − 20 = 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:

S

_{2}= 50 × 10 = 500 ftBetween 20 s to 30 s, acceleration is constant, i.e., −5 ft/s

^{2}.At 20 s, velocity is 50 ft/s.

t = 30 − 20 = 10 s

\[S_3 = ut + \frac{1}{2}a t^2 = 50 \times 10 + \frac{1}{2}\left( - 5 \right) {10}^2 \]

\[ \Rightarrow S_3 = 500 - 250 = 250 ft\]

\[ \Rightarrow S_3 = 500 - 250 = 250 ft\]

otal distance travelled is 30 s:

S

= 250 + 500 + 250

= 1000 ft

The position–time graph:

S

_{1}+ S_{2}+ S_{3}= 250 + 500 + 250

= 1000 ft

The position–time graph:

Concept: Motion in a Straight Line - Position-time Graph

Is there an error in this question or solution?

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