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

Sneha Kharya , 8 Years ago
Grade 12
anser 3 Answers
Bongi
the initial velocity is give as vi = 3i + 4j 
the acceleration is a = 0.4i + 0.3j 
and the time elapsed is t = 10s
now final velocity after 10s is vf = (3i+4j)m/s + 10s(0.4i+0.3j)m/s
vf = 7i + 7j
now speed is v =\sqrt{(7i)^2)+ (7j))} = 7\sqrt{2}
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Last Activity: 8 Years ago
SIDDHESH SHRIKANT KALEKAR
u=3i+4j. a=0.4i+0.3j. t =0By using v = u+ atWe get, v=3i+4j+(0.4i+0.3j)10 v=7i+7jSo, magnitude of v is 7*2^(0.5)=7*1.41=9.87(approx)
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Last Activity: 8 Years ago
rahul kumar
By using v=u+at ,u=3i+4j,a=0.4i+0.3j,
V=(3i+4j)+(0.4i+0.3j)*10=7i+7j
Then resultant =7√2 so,  the answer is 7√2
Last Activity: 7 Years ago
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