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Question

A particle has initial velocity (2i^+3j^) and acceleration (0.3i^+0.2j^) the magnitude of the velocity after 10s will be:


A

92units

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B

52units

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C

9units

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D

5units

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Solution

The correct option is B

52units


Step 1: Given data

Initial velocity,u=(2i^+3j^)

Acceleration, a=(0.3i^+0.2j^)

Time, t=10s

Step 2: To find

The velocity of the particle after 10s.

Step 3: Formula used

v=u+at

where v is final velocity, u is initial velocity, a is acceleration and t is the time taken.

Step 4: Calculation

v=u+at

Substituting the values in the equation,

v=(2i^+3j^)+(0.3i^+0.2j^)×10v=2i^+3j^+3i^+2j^v=5i^+5j^

Taking the magnitude of v,

v=52+52v=2×52v=52units

Therefore the magnitude of velocity of the particle after 10s is 52units.


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