Question

Can you do 4,5,7,8 please

Graph Practice Problems -v vs.t PHYS&114 Name: Use the graph below to answer questions 1-11. The graph represents the velocity of a particle as a function of time. The positive direction is right and the negative direction is left. 25 20 15 10 7 8 9 o -5 -10 -15 -20 25 time (s) 1. What is the instantaneous velocity of the particle at t-3.00s? 2. What is the instantaneous velocity of the particle at t-9.00s 3. what is the instantaneous acceleration of the particle at t=1.00s? What is the average acceleration of the particle from t=500s to t=10.0s? 5) what is the average acceleration of the particle fron t=0.00s to t-100s? 6. What is the instantaneous acceleration of the particle at t=9.00s? 7. What is the displacement of the object between t-0.00s and t-5.00s? & What is the position of the particle at t-6.00s, if it has an initial postion of 3.00m to the right of the origin? 9. At t-5.00s, is the particle speeding up, slowing down, or at rest? 10. At t-7.00s, is the particle speeding up, slowing down, or at rest? 11. At t-9.00s, is the particle speeding up, slowing down, or at rest?

Answer 1

1.

at t = 3 sec

v_{instantaneous} = 20 m/s          (since at t = 3 along x-axis , the value of y-coordinate or velocity is 20)

2.

at t = 9 sec

v_{instantaneous} = - 10 m/s          (since at t = 9 along x-axis , the value of y-coordinate or velocity is - 10)

3.

a_{instantaneous} = slope of line AB = BC/AC = (20 - 0)/(2 - 0) = 10 m/s²

4.

v_i = initial velocity at t = 5 , = 10 m/s

v_f = final velocity at t = 10 s , = 0 m/s

t = time interval = 10 - 5 = 5 sec

average acceleration is given as

average acceleration = (v_f - v_i)/t

average acceleration = (0 - 10)/5 = - 2 m/s²

5.

v_i = initial velocity at t = 0 , = 0 m/s

v_f = final velocity at t = 10 s , = 0 m/s

t = time interval = 10 - 0 = 10 sec

average acceleration is given as

average acceleration = (v_f - v_i)/t

average acceleration = (0 - 0)/10 =0 m/s²

7.

displacement = area of triangle ABH + area of triangle DIE + area of rectangle BDGH + area of rectangle IEFG

displacement = (0.5) (BH x AH) + (0.5) (DI x IE) + (BH x HG) + (IG x GF)

displacement = (0.5) ((20 - 0) x (2 - 0)) + (0.5) ((20 - 10) x (5 - 4)) + ((20 - 0) x (4 - 2)) + ((10 - 0) x (5 - 4))

displacement = 75 m

8.

displacement from t = 0 to t = 6 is given as

displacement = area of isosceles trapezium ABDJA

displacement = (0.5) (BD + AJ) (BH) = (0.5) (2 + 6) (20) = 80 m

initial position = 3 m

displacement = final position - initial position

80 = final position - 3

final position = 83 m