Stuck on basic Physics/Math Question?

You want the equation

s = ut + ½at²

s = distance moved = 1 ft (initially)

u = initial velocity = 0

a = acceleration

t = time = 1 sec

So

1 = 0 + ½ a (1²)

therefore a = 2 ft/sec²


Using the same equation with t = 2 sec.

s = 0 + ½ (2) ∙ 2²

. .= 4 ft.

In the first two seconds it has moved 4 ft, therefore in the second second it moved 4 - 1 = 3 feet.

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Ball moved 1 foot between t=0 n t=1 sec
Initial velocity V0 at t = 0 sec
Let velocity after t = 1 sec, V1
Acceleration = (V1-V0)/t, As t = 1 Acceleration = V1-V0
Let velocity after t = 2 sec, V2
As acceleration is constant
V2-V1 = V1-V0
V2+V0 = 2V1

Given, V0 = 0 and V1 = 1 as ball moved 1 foot in 1 sec
So V2 = 2 feet/sec
Acceleration 1 foot/sec²
Formula for constant acceleration is
tda = Vf² - Vi²
t - time, d - distance, a - acceleration
d = (4-1)/1*1
d = 3

So it moved 3 feet