Question

A particle moves along the x-axis. The velocity of this particle as a function of time is shown in the figure. Assume the particle is located at x = 0 m at time t = 0 s.


What is the acceleration of the particle at time t = 28.0 s? .7m/s^2


What is the position of the particle along the x-axis at time t = 28.0 s? ????


What is the net displacement of the particle between time t = 4.0 s and t = 36.0 s?
answer 2 m

Answer 1

i have attached the graph

Answer 2

i tried s=ut+1/2at^2 i got wrong answer unless i am inputting something wong ? s = -3(28)+1/2(.7)(28)^2 i got 190.40 m but it was wrong

Answer 3

For the acceleration, you just need to find the angular coefficient of the line in which contain the image of x=28s. Which is\[\Delta y/\Delta x\] = 4-(-3)/30-20 = .7m/s²

Answer 4

For the position, you need to calculate the area under the line segments, just remember that areas above the Y=0 are postive and areas below Y=0 are negative. For example: in x = 7, the area u must find is the one below the triangule with height = 4 and base = 7, so its \[4 \times 7 \div 2 \]= 14m. Can you figure the position on x=28 by yourself now?

Answer 5

had the acceleration and the displacement just having trouble getting the position of the particle along x axis at time t = 28 s

Answer 6

can anyone help ? or guide me on what to do for particle one have answers to the rest

Answer 7

go along the time axis to 28 and find the corresponding point on velocity axis thats velocity at t=28 , v= ___ the value u got
now use s=vt to get position "s"

Answer 8

v = 2.5 but it is not the right answer i am getting

Answer 9

what u got?

Answer 10

got .7 as m/s^2 acceleration when time 28s
displacement is 2m between 4 and 36s
for s=ut+1/2at^2 i got 190.40m which is wrong as i think it is wrong as acceleration is not constant !

Answer 11

i tried getting area above line to time 28 and below the line nothing seems to work

Answer 12

2.5 v x 28 t = 70m not correct either

Answer 13

Its wrong because what @ZakaullahUET is wrong, you need to consider that the velocity changes. What @Mferraciolli said is correct, you need to find the area.

Answer 14

Find the shaded area:

Answer 15

cheers will do that now !

Answer 16

Since the speed is increasing, you need to find the average speed, and since the graph is a line, that speed is the speed that divides the line in the half, and the formula is b*h/2, that is the area. That is valid for any function not only lines. Do you understand now why you need to find the area?

Answer 17

b*h/2 for the distance that you go with the average speed.

Answer 18

thank you got it was adding the wrong are underneath cheers !!!!