Saved Bookmarks
| 1. |
Hii guys.........How are you all.........I hope you all are fine..............☺️My question: Using velocity-time graph derive the relation between position-time relation? |
|
Answer» Answer: Hey mate! Here's your answer!! Consider the graph shown in the given attachment. OBVIOUSLY, the total distance s travelled by the object in time t is given by are under v-t graph. Therefore, distance travelled, s = Area OABGC Obviously, OABGC is a trapezium, whose area is (OA + BC/2) × OC. But, OA = u, BC = v and OC = t Distance travelled, s = ( \frac{oa + bc}{2} ) \TIMES oc = ( \frac{u \times v}{2} ) \times ts=( 2 oa+bc
)×oc=( 2 u×v
)×t From the velocity - time relation, we have at = v - u or t = v - u/a On SUBSTITUTING this VALUE of t in equation, we get s = ( \frac{u + v}{2} ) \times ( \frac{v - u}{a} ) = \frac{v {}^{2} - u {}^{2} }{2a}s=( 2 u+v
)×( a v−u
)= 2a v 2 −u 2 ]
➡ v² - u² = 2as Therefore, this equation is the position velocity relation for uniformly accelerated motion. Hope it helps you! ✌ ✌ ✌ |
|