Chapter 4: Q. 9 (page 106)

9. A particle moving in the x y plane $\vec{v} = (2 t \hat{i} + (3 - t^{2}) \hat{j}) m / s$
Here t is in s. What is the particle's acceleration vector at $t = 4 s$ ?

Short Answer

Expert verified

The acceleration vector of the particle at is $(2 \hat{i} - 8 \hat{j}) m / s^{2}$

Step by step solution

Step 1. Given information

The velocity vector of the particle moving in the x-y plane is $\vec{v} = (2 t \hat{i} + (3 - t^{2}) \hat{j}) m / s$ and the time instant is .

$t = 4 s$

Step 2. Explanation

The acceleration vector of the particle is,

$\vec{a} = \frac{d}{d t} (\vec{v})$

$\vec{a} = \frac{d}{d t} ((2 \hat{i} + (3 - t^{2}) \hat{j}) m / s)$

For t = 4s

The acceleration vector of the particle is

$\begin{array}{l} \vec{a} = (2 \hat{i} - 2 (4 s) \hat{j}) m / s^{2} \\ = (2 \hat{i} - 8 \hat{j}) m / s^{2} \end{array}$

Therefore, the acceleration vector of the particle is $(2 \hat{i} - 8 \hat{j}) m / s^{2}$

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9. A particle moving in the x y plane $\vec{v} = (2 t \hat{i} + (3 - t^{2}) \hat{j}) m / s$
Here t is in s. What is the particle's acceleration vector at $t = 4 s$ ?

Short Answer

Step by step solution

Step 1. Given information

Step 2. Explanation

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9. A particle moving in the x y planev→=2ti^+3−t2j^m/sHere t is in s. What is the particle's acceleration vector at t=4s?

Short Answer

Step by step solution

Step 1. Given information

Step 2. Explanation

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Most popular questions from this chapter

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9. A particle moving in the x y plane $\vec{v} = (2 t \hat{i} + (3 - t^{2}) \hat{j}) m / s$
Here t is in s. What is the particle's acceleration vector at $t = 4 s$ ?