Advertisements
Advertisements
प्रश्न
Derive the equation
S = ut+ `1/2` at2
Using a speed- time graph
Advertisements
उत्तर
The area enclosed under a velocity time curve gives the distance covered by a moving body. So total distance S covered by a uniformly accelerating body is given by area of trapezium OSQP.
S = area of trapezium OSQP.
AREA of rectangle OSRP + area of triangle PRQ.
S = OP x OS + `1/2` PR xQR.
s = u x t + `1/2` x t x at
S = ut + `1/2` at2
This is known as second equation of motion.
APPEARS IN
संबंधित प्रश्न
What conclusion can you draw about the acceleration of a body from the speed-time graph shown below .
The velocity-time graph for part of a train journey is a horizontal straight line. What does this tell you about its acceleration ?
Study the speed-time graph of a car given alongside and answer the following questions:
(i) What type of motion is represented by OA ?
(ii) What type of motion is represented by AB ?
(iii) What type of motion is represented by BC ?
(iv) What is the acceleration of car from O to A ?
(v)What is the acceleration of car from A to B ?
(vi) What is the retardation of car from B to C ?
What type of motion is represented by the following graph ?
From the displacement-time graph of a cyclist given below in the Figure, find The time after which he reaches the starting point .
Draw displacement – time graph for the following situation:
When a body is stationary.
The area of the right triangle under a speed-time graph is 500 m, in a time interval of 20 s. What is the speed of the body? Is the motion uniform or non-uniform?
Draw the following graph:
Distance versus time for a body at rest.
Interpret the following graph:
The area under the v-t graph represents a physical quantity that has the unit.
