A freely falling body describes a distance x in first 2 seconds and a diatance y in next 2 seconds .Then show that y=3x
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Answered by
38
Object is falling freely
In first two seconds it will cover distance of
d = ut + [(at²)/2]
Here, d = x , u = 0, t = 2s, a = g = 9.8 m/s²
x = (0)(2) + [(9.8 * 2²)/2]
x = 0 + (9.8 * 4)/2
x = 19.6 m ---------------------(1)
So in first two seconds it will cover 19.6 m distance
For first 2 seconds that is till it covers x meter distance it's velocity at that point will be
v = u + at
Here, a = g = 9.8 m/s² , t = 2s , u = 0 m/s
v = 0 + (9.8 * 2)
v = 19.6 m/s
So, while it start for next two seconds its initial velocity will be 19.6 m/s
Now, we have to find y that is distance covers in next 2 seconds will be
d = ut + [(at²)/2]
Here, d = y , u = 19.6 m/s, t = 2s, a = g = 9.8 m/s²
y = (19.6)(2) + [(9.8 * 2²)/2]
y = 39.2 + (9.8 * 4)/2
y = 58.8 m -----------------(2)
So it will cover y = 58.8 m distance in next 2 seconds.
Now, Dividing equation 2 with equation 1
![\frac{y}{x} = \frac{58.8}{19.6} \frac{y}{x} = \frac{58.8}{19.6}](https://tex.z-dn.net/?f=+%5Cfrac%7By%7D%7Bx%7D++%3D++%5Cfrac%7B58.8%7D%7B19.6%7D+)
![\frac{y}{x} = 3 \frac{y}{x} = 3](https://tex.z-dn.net/?f=+%5Cfrac%7By%7D%7Bx%7D++%3D+3)
![y = 3x y = 3x](https://tex.z-dn.net/?f=y+%3D+3x)
Hence proved
In first two seconds it will cover distance of
d = ut + [(at²)/2]
Here, d = x , u = 0, t = 2s, a = g = 9.8 m/s²
x = (0)(2) + [(9.8 * 2²)/2]
x = 0 + (9.8 * 4)/2
x = 19.6 m ---------------------(1)
So in first two seconds it will cover 19.6 m distance
For first 2 seconds that is till it covers x meter distance it's velocity at that point will be
v = u + at
Here, a = g = 9.8 m/s² , t = 2s , u = 0 m/s
v = 0 + (9.8 * 2)
v = 19.6 m/s
So, while it start for next two seconds its initial velocity will be 19.6 m/s
Now, we have to find y that is distance covers in next 2 seconds will be
d = ut + [(at²)/2]
Here, d = y , u = 19.6 m/s, t = 2s, a = g = 9.8 m/s²
y = (19.6)(2) + [(9.8 * 2²)/2]
y = 39.2 + (9.8 * 4)/2
y = 58.8 m -----------------(2)
So it will cover y = 58.8 m distance in next 2 seconds.
Now, Dividing equation 2 with equation 1
Hence proved
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Answered by
17
Initial velocity u=0(Bz it is at rest)
Distance travelled in first two seconds is
S = ut +1/2at^2
X = (0*2) + 1/2a(2)^2
X = 0 + 1/2a(4)
X = 0+2a
X = 2a ----->1st equation
Distance travelled in first 4 seconds is
S = ut + 1/2at^2
X + Y = (0*4) + 1/2a(4)^2
X+ Y = 0 + 1/2a(16)
X + Y = 0 + 8a
X + Y = 4(2a) (bz frm equation i.e x =2a)
X + Y = 4X
Y =4X - X
Y =3X
Hence proved
Sorry as I couldn't write square of t so I used t^2
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