The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Average speed and position

An object moves in a straight line, its position as a function of time is given by the equation: x= -5 + 20t - 2.5 t^2
The object's average speed between t=0 and t=6 s is:

I know it's not right to just plug in t= 0s and t= 6s. Then how I can do it? How I can know the real total distance traveled?
Original post by Daniel Atieh
An object moves in a straight line, its position as a function of time is given by the equation: x= -5 + 20t - 2.5 t^2
The object's average speed between t=0 and t=6 s is:

I know it's not right to just plug in t= 0s and t= 6s. Then how I can do it? How I can know the real total distance traveled?

Wait so are you asked to find the total distance travelled or the speed?
Original post by langlitz
Wait so are you asked to find the total distance travelled or the speed?

Well, they need the average speed, so this means I will need to find the total distance traveled right?
Original post by Daniel Atieh
Well, they need the average speed, so this means I will need to find the total distance traveled right?


Then yes you do just plug in t=0 and t=6 to find the distance travelled then divide by 6
Original post by langlitz
Then yes you do just plug in t=0 and t=6 to find the distance travelled then divide by 6

That's what I also thought, but there's someone who solved it in my paper ( A teacher ), and he said that there's a reflection point, he differentiated it, and found t=4s. I don't know what's this :/
Original post by Daniel Atieh
That's what I also thought, but there's someone who solved it in my paper ( A teacher ), and he said that there's a reflection point, he differentiated it, and found t=4s. I don't know what's this :/


You actually get the same answer either way

x=−5+20t−2.5t2 x= -5 + 20t - 2.5 t^2
dx/dt=20−5tdx/dt=20-5t
t=4t=4 when dx/dt = 0

x=−5+20(4)−2.5(4)2=−2+80−40=35m x = -5+20(4)-2.5(4)^2 = -2+80-40 = 35 m


OR

at t= 0 x=−5+20(0)−2.5(0)2=−5x=-5+20(0)-2.5(0)^2=-5
at t=6 x=−5+20(6)−2.5(6)2=25mx=-5+20(6)-2.5(6)^2= 25 m

Thus change in x is x = 25-(-5)= 35 m
Original post by langlitz
You actually get the same answer either way

x=−5+20t−2.5t2 x= -5 + 20t - 2.5 t^2
dx/dt=20−5tdx/dt=20-5t
t=4t=4 when dx/dt = 0

x=−5+20(4)−2.5(4)2=−2+80−40=35m x = -5+20(4)-2.5(4)^2 = -2+80-40 = 35 m


OR

at t= 0 x=−5+20(0)−2.5(0)2=−5x=-5+20(0)-2.5(0)^2=-5
at t=6 x=−5+20(6)−2.5(6)2=25mx=-5+20(6)-2.5(6)^2= 25 m

Thus change in x is x = 25-(-5)= 35 m

Many thanks. Now I can use this for the average velocity, as the total distance is 50 m. How do we arrive to it?
Original post by langlitz
You actually get the same answer either way

x=−5+20t−2.5t2 x= -5 + 20t - 2.5 t^2
dx/dt=20−5tdx/dt=20-5t
t=4t=4 when dx/dt = 0

x=−5+20(4)−2.5(4)2=−2+80−40=35m x = -5+20(4)-2.5(4)^2 = -2+80-40 = 35 m


OR

at t= 0 x=−5+20(0)−2.5(0)2=−5x=-5+20(0)-2.5(0)^2=-5
at t=6 x=−5+20(6)−2.5(6)2=25mx=-5+20(6)-2.5(6)^2= 25 m

Thus change in x is x = 25-(-5)= 35 m

Sorry for appearing again after ages. Gotta ask about the total distance traveled. So I should add 5 to the 35 m you found by differentiating, right?
Original post by Daniel Atieh
Sorry for appearing again after ages. Gotta ask about the total distance traveled. So I should add 5 to the 35 m you found by differentiating, right?

Why would you add on 5?

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you