Question

A particle starts moving at $t = 0$ along $x$ -axis from $60 m s^{- 1}$ , with an initial velocity $10 m s^{- 1}$ . The acceleration of the particle is linearly increasing as shown by its acceleration-time graph. Find the change in velocity of the particle during the time interval $t = 5 s$ to $10 s$ . Also, find the final velocity of the particle at the time $t = 10 s$ .

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Accepted Answer

Change in velocity from time $t_{1}$ and $t_{2}$ is related to acceleration $a$ as

$∆ v = \int_{t_{1}}^{t_{2}} a d t$ .

This can be calculated as the area under $a - t$ graph.

The change in velocity of the particle during the time interval $t = 5 s$ to $t = 10 s$ .

$=$ Area of the graph between $t = 5 s$ and $t = 10 s$

$\Rightarrow Δ v_{1} = \frac{(6 + 7)}{2} \times (10 - 5) = \frac{13}{2} \times 5 = \frac{65}{2} m s^{- 1}$ .

The change in velocity of the particle during the time interval $t = 0$ to $t = 10 s$

$=$ Area of the graph between $t = 0$ and $t = 10 s$

$Δ v_{2} = \frac{(5 + 7)}{2} \times (10 - 0) = \frac{12}{2} \times 10 = 60 m s^{- 1}$ .

The velocity of the particle at $t = 0$ is $v_{0} = 10 m s^{- 1}$ and change in velocity during this interval is $60 m s^{- 1}$ .

Hence, the velocity of the particle at $t = 10 s$ ,

$v_{2} = v_{0} + ∆ v_{2} = 10 + 60 = 70 m s^{- 1}$ .

B M Sharma Solutions for Chapter: Basic Mathematics, Exercise 69: ILLUSTRATION 2.63

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Basic Mathematics, Exercise 69: ILLUSTRATION 2.63

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