Advertisements
Advertisements
प्रश्न
A train starting from rest picks up a speed of 20 ms−1 in 200 s. It continues to move at the same rate for the next 500 s and is then brought to rest in another 100 s.
- Plot a speed-time graph.
- From graph calculate
(a) uniform rate of acceleration
(b) uniform rate of retardation
(c) total distance covered before stopping
(d) average speed.
Advertisements
उत्तर
(i)
(ii)
(a) Uniform rate of acceleration = Slope of v.t. graph from O to A.
= `"AE"/"OE"=20/200` = 0.1 ms−2
(b) Uniform rate of retardation = Slope of v.t. graph from BC
= `"BD"/"CD"=20/100` = 0.2 ms−2
(c) Total distance covered before stopping
= ar(ΔOAE) + ar(ΔABDE) + ar(ΔBCD)
= `1/2xx"OE"xx"AE"+"AE"xx"ED" + 1/2xx"BD"xx"DC"`
= `1/2xx200xx20+20xx500+1/2xx20xx100`
= 2000 + 10000 + 1000
= 13000 m
= 13 km
(d) Average speed = `"Total distance covered"/"Total time taken"=13000/800`
= `130/8`
= 16.25 ms−1
APPEARS IN
संबंधित प्रश्न
What type of motion is represented by the following graph ?
Diagram shows a velocity – time graph for a car starting from rest. The graph has three sections AB, BC, and CD.
Is the magnitude of acceleration higher or lower than that of retardation? Give a reason.
Interpret the following graph:
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 parallel to line axis?
Its time-displacement graph is a straight line.
Derive the equation
S = ut+ `1/2` at2
Using a speed- time graph
Figure shows the distance-time graph of three students A, B and C. On the basis of the graph, answer the following :
Which of the three is traveling the fastest?
A straight line parallel to the x-axis in the velocity-time graph represents the object moving in ______.
An object is moving in a positive direction with positive acceleration. The velocity-time graph with constant acceleration which represents the above situation is:
