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Q.

A particle has an initial velocity 3i^+4j^ and an acceleration of  0.4i^+0.3j^. Its speed after 10s is

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a

7 units

b

72 units 

c

10 units 

d

8.5 units

answer is B.

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Detailed Solution

v¯=u¯+at-

v¯=(3i^+4j^)+(0.4i^+0.3j^)10

=7i^+7j^=72 unit 

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