The position of the particle is given by x=2t2−4t+4 in m. The average speed of the particle in 6sec is
A
436m/s
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B
263m/s
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C
273m/s
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D
233m/s
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Solution
The correct option is B263m/s As we know that the velocity of the particle as a function of time is given by dxdt=d(2t2−4t+4)dt ⇒v=4t−4
Now, v(t)=4t−4, we will check if the particle has changed its direction or not within the asked time interval, i.e. the time when velocity of the particle becomes zero. ⇒4t−4=0⇒t=1sec
And, we have x(t)=2t2−4t+4
So, at t=0,x=4m,
at t=1,x=2×12−4×1+4=2m
at t=6,x=2×62−4×6+4=52m
The motion of the particle can be represented as shown
So, the average speed of the particle is 2×2+486=263m/s