Total Distance with Acceleration and Deceleration Phases

❓ Question

A car accelerates from rest at a constant rate $a$ for some time after which it decelerates at a constant rate $b$ to come to rest. If the total time elapsed is $t$ seconds, then the total distance travelled is ________.

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🖼 Question Image

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✍️ Short Explanation

This is a motion graph based kinematics problem.

👉 Car first accelerates
👉 Then decelerates to rest
👉 Use velocity-time relation and area under graph.

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🔷 Step 1 — Let Maximum Velocity be $v$ 💯

Acceleration phase:

$$v=at_1$$

So:

$$t_1=\frac{v}{a}$$

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During retardation:

$$v=bt_2$$

So:

$$t_2=\frac{v}{b}$$

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🔷 Step 2 — Total Time Relation

Given total time:

$$t=t_1+t_2$$

Substitute values:

$$t=\frac{v}{a}+\frac{v}{b}$$
$$t=v\left(\frac1a+\frac1b\right)$$
$$t=v\left(\frac{a+b}{ab}\right)$$

Thus:

$$v=\frac{abt}{a+b}$$

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🔷 Step 3 — Distance Using Velocity-Time Graph

Velocity-time graph forms a triangle.

Total distance:

$$S=\frac12 \times \text{base} \times \text{height}$$

Base:

$$=t$$

Height:

$$=v$$

So:

$$S=\frac12 vt$$

Substitute $v$:

$$S=\frac12 \times \frac{abt}{a+b}\times t$$
$$S=\frac{ab t^2}{2(a+b)}$$

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🔷 Step 4 — JEE Trap Alert 🚨

❌ Separate distances unnecessarily calculate karna

❌ Velocity-time graph use na karna

❌ Deceleration ko negative sign mein confuse ho jaana

Remember:

$$\text{Distance = Area under } v-t \text{ graph}$$

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✅ Final Answer

$$\boxed{ \frac{ab t^2}{2(a+b)} }$$

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