A block is fired up an incline and a record of its position as a function of tim

Question

A block is fired up an incline and a record of its position as a function of time is shown in the graph. The y-coordinate in the graph represents the distance along the incline, measured from the base of the incline.

Coordinates for the data points are as shown in the table below.

A tangent line has been drawn at t = 1.00 s, and this line also passes through the point (2.00 s, 21.00 m). A second tangent line has been drawn at t = 4.00 s, and this line also passes through the point (4.50 s, 0 m).

Determine the following for the block. (Indicate the direction with the sign of your answer.)

(a) the average velocity as it travels from the launch position to the maximum height
__ m/s

(b) the instantaneous velocity at a time equal to half the time to go from the launch position to the maximum height
__ m/s

(c) the instantaneous velocity at the top of the trajectory
__ m/s

(d) the instantaneous velocity when it is back down to the launch position
__ m/s

(e) the launch velocity
__ m/s

Time t(s) 0 1.00 2.00 3.00 4.00 Position y(m) 6.00 15.00 18.00 15.00 6.00 y (m) 20 15 10 5 1 L t (s) 5 2 3 4

Explanation / Answer

from the givne graph and data

a. as the block travels from launch position to maximum height

the datapoints are

0 6

1 15

2 18

3 15

4 6

the equation from regression we get is

y = -3t^2 + 12t + 6

y' = -6t + 12

y" = -6 m/s/s

so, time taken to reach maximum height = t'

at t', dy/dt = 0

t' = 2 s

so, vav= (18-6)/2 = 6 m/s

b. at t = 1 s

y' = -6 + 12 = 6 m/s

c. at top of trajectory instantaneous velocity is 0 m/s as y' = 0

d. at the launch position

t = 4 s

y' = -12m/s

e. y' at t= 0 is 12 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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