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Q. The position of an object moving along x-axis is given by x= a + bt2 where a=8.5m, b=2.5m/s2 and t is measured in seconds. What is its velocity at t=0s and t=2.0s. What is the average velocity b/w t=2.0s and t= 4.0s ?pls reply soon

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 we know that dx/dt = velocity..on differentiating the equation we get v=2bt(using dx^n/dx=nx^(n-1) and d constant/dt = 0)

at t=0s v=0m/s and at t= 2s, v=10m/s

at t=2s v1=10m/s and at t=4s v2=20 so average is (10 +20 )/2 = 15

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how did u get v=2bt? pls do derive it. And wat formula hve u used for avg.velocity?? out der . coz avg.velocity= x2-x1/ t2-t1. y is it 10+ 20/ 2 ???

  • -8

srry d avg. velocity = x2-x1/ t2-t1

  • 5

Velocity= dx / dt.

So, Differenciating x we get -

v = d( a + bt2 ) / dt.

v = 2bt ( On Differentiation xn = nxn-1 ).

Velocity at t=0,

v = 2 X 2.5 X 0.

v = 0.

Velocity at t-2,

v = 2 X 2.5 X 2.

v = 10 m/s.

v2 at t=4s,

v2 = 8.5 + 2.5 (2 X 2)

v2 = 18.5 m/s.

v1 = 8.5 + 2.5 (4 X 4)

v1= 48.5 m/s.

Average Velociy= v2 - v1 / t2 -t1.

                             = 48.5 - 18.5 / 4 - 2.

                             = 30 / 2.

                             = 15m/s.

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