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CBSE Class 9 Answered

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Asked by Kanhaiya | 16 Sep, 2015, 07:57: PM
answered-by-expert Expert Answer
(a) Between A and Bbus has maximum speed i.e. 20 m/s. Hence, Between A nd B of the journey the bus was moving fastest.
(b) During C and the bus was at rest.
(c) The acceleration between the points D and E is nothing but, acceleration of the body in time interval 10 s to 44 s
    begin mathsize 12px style equals space fraction numerator 0 minus 12 over denominator 44 minus 40 end fraction space equals space fraction numerator negative 12 over denominator 4 end fraction space equals space 3 space straight m divided by straight s squared end style
(d) The graph BC represents the retardation.
(e) The distance travelled will be equal to the area under the graph from 0 to 28 seconds.
     Area of the trapezium OABC = begin mathsize 12px style fraction numerator AB space plus space OC over denominator 0 end fraction space cross times space straight A space to space point space at space 8 space straight s space equals space fraction numerator 12 space plus space 28 over denominator 2 end fraction space cross times space 20 space equals space 160 space straight m space end style
Answered by Priyanka Kumbhar | 18 Sep, 2015, 11:07: AM

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CBSE 9 - Physics
Time- 0 10 20 30 40 50 Speed- 5 10 15 20 25 30 Find the distance travelled in 50 seconds
Asked by akshatdhamija1409 | 15 Jul, 2022, 11:16: PM
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Starting from rest, an object is uniformly accelerated to a speed of 60m/s in 20s. The object travels with this speed for next 40s and is then brought to rest by a uniform retardation in the next 30s.Sketch velocity time graph and calculate the acceleration, retardation and the total distance travelled. Hint: acceleration =rate of change of velocity.  distance covered =area under velocity-time graph .
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CBSE 9 - Physics
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Asked by Santosh.kumar0317 | 25 Aug, 2019, 02:15: PM
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