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 = ba/2
= 33.311.1/2
= 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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