Solve this-5- nbsp-The position Of an object- moving in one nbsp-dimension- is given by the formula nbsp-x-t- - -4-2t2 - 2-6-m- Calculate its -i- average nbsp-velocity in the time interval from t - 0 to t - 3s nbsp-and -ii- Instantaneous velocity at t - 3 s- nbsp-d-xn-dt -160- -160-nxn-1 -160-

Dear Student , Here in this case the average velocity can be derived as : x=4.2t2+2.6 #8658;dxdt=v=8.4tNow #160;velocity #160;at #160;time #160;t=3 ...

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Byju's Answer

Standard XII

Physics

Relative Velocity

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Question

Solve this:
5) The position Of an object, moving in one
dimension, is given by the formula
x(t) = (4.2t²+ 2.6)m. Calculate its (i) average
velocity in the time interval from t = 0 to t = 3s
and (ii) Instantaneous velocity at t = 3 s.
$[\frac{d (x^{n})}{d t} = n x^{n - 1}]$

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Solution

Dear Student ,
Here in this case the average velocity can be derived as :

$x = 4.2 t^{2} + 2.6 \Rightarrow \frac{d x}{d t} = v = 8.4 t N o w v e l o c i t y a t t i m e t = 3 s i s, v_{1} = 8.4 t = 25.2 m / s N o w v e l o c i t y a t t i m e t = 0 i s, v_{2} = 0 m / s S o t h e a v e r a g e v e l o c i t y i s, V_{a v e r a g e} = \frac{v_{1} + v_{2}}{2} = 12.6 m / s N o w t h e i n s \tan \tan e o u s v e l o c i t y a t t i m e t = 3 s i s, v' = 8.4 \times 3 = 25.2 m / s$
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Solve this: 5) The position Of an object, moving in one dimension, is given by the formula x(t) = (4.2t2 + 2.6)m. Calculate its (i) average velocity in the time interval from t = 0 to t = 3s and (ii) Instantaneous velocity at t = 3 s. d(xn)dt = nxn-1

Solve this:
5) The position Of an object, moving in one
dimension, is given by the formula
x(t) = (4.2t²+ 2.6)m. Calculate its (i) average
velocity in the time interval from t = 0 to t = 3s
and (ii) Instantaneous velocity at t = 3 s.
$[\frac{d (x^{n})}{d t} = n x^{n - 1}]$