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Q.

A particle has an initial velocity of 3i^+4j^ and an acceleration of  4i^+3j^. Then its speed in 10s is…........units.

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answer is 54.82.

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Detailed Solution

Given initial velocity  u¯=3i^+4j^
Acceleration  a¯=4i^+3j^
Time t=10s.
 from equation  V¯=u¯+a¯t
V¯=(3i^+4j^)+(4i^+3j^)10=(3i^+4j^)+(40i^+30j^)=43i^+34j^  
 Magnitude of velocity (or) speed
V=432+(34)2=1849+1156=3005=54.82 units

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