ICSE Class 9 Answered

PLS NEED A QUICK SOLUTION FOR THE ATTACHED QUESTION

Asked by shyluraja21 | 04 Oct, 2021, 16:32: PM

Expert Answer

1. According to the corresponding distance - time values we can see that the car travels equal distances in equal intervals of time. Thus, the car is moving in uniform motion.

The car is thus moving with constant speed = distance/ time = 20/10 = 2 m/s

2. According to the given velocity - time values, the v-t graph will be as follows: -

i) According to this v-t graph, we can see that velocity goes on increasing uniformly. Thus, the bus is moving with uniform acceleration.

ii) The initial velocity of bus is 5 m/s

iii) Acceleration of bus during the time interval t =0s to t = 25 s is,

A c c e l e r a t i o n space equals space fraction numerator v f space minus space v i over denominator t f space minus space t i end fraction space equals space fraction numerator 30 minus 5 over denominator 25 minus 0 end fraction equals 25 over 25 equals 1 space m divided by s squared

iv) Distance travelled = Area under curve of v-t

or we can calculate the same using the thrid equation of motion

v² - u² = 2as

v = 30 m/s, u = 5m/s

a = 1 m/s²

Thus,

30² - 5² = 2 (1) x s

s = 875/2 = 437.5 m

Answered by Shiwani Sawant | 04 Oct, 2021, 20:03: PM

ICSE Class 9 Answered

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ICSE Class 9 Answered

Application Videos

velocity-time graph

speed time graph

Concept Videos

Motion in One Dimension

Motion

Motion

Motion

Motion

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