A car starts from X = 11m at t = 0 and moves with the velocity graph shown in figure below. [email protected] 2s,3s,4s?

Question

A car starts from a position of ( x = 11 \, 	ext{m} ) at ( t = 0 ) and moves according to the velocity graph shown below. 
GRAPH: [Insert link here]
A) What is the object's position at ( t = 2\, 	ext{s} )?
B) What is the object's position at ( t = 3\, 	ext{s} )?
C) What is the object's position at ( t = 4\, 	ext{s} )?  
Thank you! 🙂

Anonymous · Accepted Answer

The position change is the area under the velocity vs time curve

Don’t forget to include the initial position (x = 11m @ t = 0)

A) So x = x0 + area under curve from 0 to 2 seconds

x = 11 + 1/2b(h1 + h2) (area of a t ɾąքҽ zoid) = 11 + 1/22(12 + 4) = 27m

B) now the area is a triangle so x = 11 + 1/2123 = 11 + 18 = 29m

C) now there is a negative area so x = 11 + 1/2123 – 1/214 = 27m

Anonymous · Answer

the object is slowing down, after 3 seconds it starts to go backwards, velocity = displacement/time therefore displacement = velocity x time
at t=2 its at 8m because velocity=4 time=2 —->4×2=8
at t=3 its at 0m because velocity=0 time=3 —->0x3=0
and finally t=4 is -16 because velocity=-4 and time=4 therefore its -16 so its 16 from where it started
Plus the guy belows method works to (y)

Daron Rolfson · Answer

the link is not workin.! ?
well.., a hint dat i can give u is..  try calculating the area under the velocity- time graph ( in case it is one).. that will give u THE CHANGE IN POSITION.!

Anonymous

A car starts from X = 11m at t = 0 and moves with the velocity graph shown in figure below. [email protected] 2s,3s,4s?

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