an object moving with initial velocity u along a straight line attains maximum velocity v after travelling a distance s. assume the acceleration is uniform. derive the relationship between position and velocity of the object by graphical method.
Answers
Answer:
third third equation of motion ( 2as=v -u)
(position velocity relation)
distance (s)=area under the graph
=area of trapezium o a b c
1/2 *h*(sum sum of 11 sides)
use the following attachment to solve the question....
Answer:
The relation will be v² - u² = 2as between position and velocity of the object.
Explanation:
Refer to the diagram attached for more clarity.
AB is velocity-time graph. From point B, draw a ⊥ on x and y axis. From the graph we know that,
initial velocity = u = OA = CD
final velocity = v = OE = BC
time = t = OC = AD
Distance (S) = Area under the graph
S = Area of trapezium OABC
S = (1/2)×OC×(OA+BC)
S = (1/2)×t×(u+v)
We know : [v=u=at]
Also we know : [(v-u)/a=t]
S = (1/2)×(v-u)/a×(v+u)
S = 1/(2a)×(v)²-(u)²
2aS = (v)²-(u)²
Hence the relation between position and velocity of the object can be expressed by 2aS = (v)²-(u)², where s is the displacement , a is acceleration, u is initial velocity, v is final velocity.
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