I need to see the formula and the step Thank you A position-time graph for a par

Question

I need to see the formula and the step Thank you A position-time graph for a particle moving along the x axis is shown in the figure below. x (m) 12 10 123 5-6 t (s) (a) Find the average velocity in the time interval t . 1.00 s to t-3.50 s. (Indicate the direction with the sign of your answer.) m/s (b) Determine the instantaneous velocity at t- 2 s (where the tangent line touches the curve) by measuring the slope of the tangent line shown in the graph. (Indicate the direction with the sign of your answer.) m/s (c) At what value of t is the velocity zero?

Explanation / Answer

(a)average velocity = distance covered in duration / duration

where duration = 3.5-1 = 2.5s

for the above curve which is parabolic, distance covered will be area covered under the curve = integral of xdt from 1 to 3.5

here, parabola is of the form x2=4ay, but is not in the direct form. By applying the corrections, the formula is (x-4)2=4a(y-2)

putting values of any point , say, 6,5 , a is obtained as 1/3

so, formula of this curve after applying the corrected axes i.e. x in place of y and t in place of x, is (t-4)2=4(x-2)/3

so, x= 3(t-4)2/4 + 2

now, evaluate the integral using mathematical tools, area under the curve is integral[(3(t-4)2/4 + 2))dt] from 1 to 3.5

or, area =11.718 units or 11.72 m

average velocity = 11.72/2.5 = 4.68m/s

(b)instantaneous velocity from the graph and the directions given,and taking 2 points on the line, is,12-2 / 3-0.5 = 4m/s

(c)velocity is zero where the above formula, or the line drawn as above, for such point gives zero value. This happens at t = 4s.

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