The velocity - time graph of a body moving in a straight line is shown in figure. The displacement and distance travelled by the in 6 seconds are respectively :
Figure. Nao.
This is the quest.
v=at d=v(initial)+1/2 at^2 d=vt
Displacement is movement, whether forward or backwards.
distance is foward movement.
for displacement, find the 'net' area under the v-t curve. for distance, find the magnitude of areas of each portion, and add 'em up.
if you are taking calc, do waht @apoorvk said,
otherwise, you might want to do it the elementary way, even though it is in essence the same thing.
Basically, since distance=velocity*time, we can add up each of these segments of constant velocity.
Now, for distance, remember that the velocity is negative sometimes and if that is the case, subtract. If, however, you are calculating displacement, calcuate the absolute value of the distance traveled in each segment, and add them up.
i can not still understand any easy method ? @apoorvk please explain in hindi it would be easier for me
can u tell me through formula putting i mean which formula to use here ? etc.
First, we need to know the difference between distance and displacement. Do you know?
(This is important.)
yes distance is a scalar displacement is a vector quantity
Ok, good.
Now, tell me if acceleration is constant in each segment, whether it is variable, or whether there is no acceleration.
NO ACCELERATION
^correct!
Good. Since there is no acceleration, let's not worry about it.
Do you know that s=vt ?
distance=velocity*time. Find the distance that is made by each of those three segments, add them up, and there is your resultfor distance.
ok s = ut + 1/2 (at2) = , since no acceleration is there hence s=ut
Tell me what you get.
8m = displacement and 16m = distance hey inkywood thanks a lot can u add me here https://www.facebook.com/kushashwaravi.shrimali?ref=tn_tnmn please
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Kay, I'm maru hedg ehog.
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i accpeted thanks a lot bbye
Wait, flip those around. Displacement is 16, and distance is 8
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