Posted by **armando** on Thursday, August 25, 2011 at 6:34pm.

a body moving with a uniform acceleration travels distances of 24m and 64m during the first two equal consecutive intervals of time each of duration 4s.determine the initial velocity and acceleration of moving body.

- physics -
**Damon**, Thursday, August 25, 2011 at 6:48pm
d = vi t + (1/2) a t^2

24 = vi (4) + (1/2) a (4)^2

88 = vi (8) + (1/2) a (8^2)

--------------------------------

48 = 8 vi + 16 a

88 = 8 vi + 32 a

----------------- subtract

-40 = -16 a

a = 2.5 m/s^2

24 = 4 vi + 20

vi = 1 m/s

check my arithmetc

- physics -
**Anonymous**, Tuesday, May 21, 2013 at 8:27am
a motor cyclist goes from station a to b at 40km/h and then returns from b to a 120km/h.calculate average speed,average velocity during the round trip

- physics -
**Anonymous**, Wednesday, June 10, 2015 at 11:51am
case 1:

from a to b velocity=40km/h=11.1m/s

let 11.1=a

case 2:

from b to a velocity=120kh/h=33.3m/s

let 33.3=b

avg.speed = b-a/2

= 33.3-11.1/2

= 22/2

= 11m/s

avg. speed= 11m/s=40km/h

avg.velocity=0

because avg. velocity is total displacement/time.

now the cyclist travels from a to b then from b to a, back to its original

position. so the avg.velocity is zero.

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