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Chapter 4: Q. 9 (page 106)

9. A particle moving in the x y planev→=2ti^+3−t2j^m/s

Here t is in s. What is the particle's acceleration vector at t=4s?

Short Answer

Expert verified

The acceleration vector of the particle at is(2i^−8j^)m/s2

Step by step solution

01

Step 1. Given information

The velocity vector of the particle moving in the x-y plane is v→=2ti^+3−t2j^m/sand the time instant is .

t=4s

02

Step 2. Explanation

The acceleration vector of the particle is,

a→=ddt(v→)

a→=ddt2i^+3−t2j^m/s

For t = 4s

The acceleration vector of the particle is

a→=(2i^−2(4s)j^)m/s2=(2i^−8j^)m/s2

Therefore, the acceleration vector of the particle is(2i^−8j^)m/s2

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