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
-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
from a to b velocity=40km/h=11.1m/s
from b to a velocity=120kh/h=33.3m/s
avg.speed = b-a/2
avg. speed= 11m/s=40km/h
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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