Question

A particle covers 25 cm and 33 cm in the 5 th and 7 th
seconds, respectively. What is the velocity of the parti-
cle 9 seconds after the initiation of the journey?
[ 43 cm.s-1​

Answer 1

Answer:

33- 25 = 8

In 2 second change in velocity is 8 cm/ s ( distance/ time is velocity)

Acceleration = 4 cm / s^2

Distance travelled in the 5 th second = 25 = u+ (5+4)a/2

Hence u =7 cm/s

V at the end of 9 s = 7 + 4*9 = 43 cm/s

Explanation:

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Answer 2

Explanation:

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