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.

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

Bro from where did the 88 come it is 64

Answer 5

Sum of both the distance (24+64)