Suppose that a quadratic fit to a position plot (see section 3.4.3.1) yields the

Question

Suppose that a quadratic fit to a position plot (see section 3.4.3.1) yields the following fit result: y = 4.97t^2 + 1.53t + 0.0503. From this fit result: A) Determine the acceleration due to gravity of the particle. B) Determine the initial speed of the particle. Plot the position of a panicle (see Equation (3.2)) as a function of time for an object from rest for the time of 0 seconds to 1.2 seconds in time intervals of 0.2 seconds (a total of seven points). Plot the speed (see Equation (3.1)) of the same particle as in problem number 2 above for the same time intervals. Suppose that the acceleration due to gravity was measured in a free fall experiment to be (9.777 plusminus 0.252) 3/s^2. Does this value agree within the uncertainty with the given value for the acceleration due to gravity in Mobile. Alabama? (Show your work. Don t just answer 'yes' or 'no")

Explanation / Answer

A]

y = 4.97t2 + 1.53t + 0.0503

v = dy/dt

so differentiate the above expression to get velocity

v = 2(4.97)t + 1.53 + 0

and a = g = dv/dt

so differentiate the velocity expression to give

a = 2(4.97) + 0 + 0 = 9.94 m/s2.

B]

v = 9.94t + 1.53

so, at t = 0s, v(0) = 9.94(0) + 1.53 = 1.53 m/s. This is the initial velocity.

2] Since initial velocity is zero, the height equation will be:

y = ut + (1/2)at2 = 0 + (1/2)gt2

the curve will be an upward parabola for t > 0s

Plot the curve for time spacing of 0.2s. So, for t = 0s to t = 1.2s, this would correspond to plotting 7 points on the graph.

3] v = u + at = 0 + gt

This is an increasing linear function of time and hence will be a straight line of slope m = g.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.