A particle starts from x=0 and moves along a straight line such that its velocity v depends on position x as v=x+4. For the first four seconds, average velocity of the particle (in m/s) is
A
e4+1
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B
e4−1
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C
e4+4
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D
e4−4
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Solution
The correct option is Be4−1 Given, v=x+4 dxdt=x+4
Integrating on both sides ∫dxx+4=∫dt ln(x+4)=t+C x+4=e(t+C)
At t=0,x=0 ⇒x=4et−4 v=4et>0fort>0 ⇒Total distance=Total displacement
∴ Average velocity for first 4 seconds =Total displacementTotal time =4e4−44=e4−1