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
Draw velocity – time graph for the following situation:
When a body is moving with variable velocity, but uniform acceleration.
A body at rest is thrown downward from the top of the tower. Draw a distance – time graph of its free fall under gravity during the first 3 seconds. Show your table of values starting t = 0 with an interval of 1 second, (g = 10 ms−2).
From the diagram given below, calculate distance covered by body.
Represent the location of an object described as at 15 m, 45o west of north, on a graph paper.
Given on th e side are a few speed - time graphs for various objects moving along a stra ight line. Refer below figure. (a), (b), (c) and (d).
Which of these graphs represent
(a) Uni form motion
(b) Motion with speed increasing
(c) Motion with speed decreasing and
(d) Motion with speed oscillating.?
Draw the following graph:
Speed versus time for a non-uniform acceleration.
Interpret the following graph:
What can you say about the nature of motion of a body of its displacement-time graph is:
A straight line inclined to the time axis with an acute angle?
Draw distance-time graph to show:
Uniform velocity
Draw distance-time graph to show:
Decreasing velocity
