1) A particle, initially at rest starts moving with a uniform acceleration 'a'. The ratio of the distances covered by it in first second and the first 3s is... (please explain and give the correct answer) …
The formulas which you can use are:
[tex]V_{avg} = u+v/2[/tex]
[tex]s = [\frac{u+v}{2} ]t[/tex]
[tex]s = ut+\frac{1}{2}at^{2}[/tex]
[tex]s = vt-\frac{1}{2}at^{2}[/tex]
[tex]t = \frac{v-u}{a}[/tex]
[tex]v^{2} -u^{2} = 2as[/tex]
s = displacement
a = acceleration
u = initial velocity
v = final velocity
t = time
The formulas which you can use are:
[tex]V_{avg} = u+v/2[/tex]
[tex]s = [\frac{u+v}{2} ]t[/tex]
[tex]s = ut+\frac{1}{2}at^{2}[/tex]
[tex]s = vt-\frac{1}{2}at^{2}[/tex]
[tex]t = \frac{v-u}{a}[/tex]
[tex]v^{2} -u^{2} = 2as[/tex]
s = displacement
a = acceleration
u = initial velocity
v = final velocity
t = time
Answers & Comments
Verified answer
Answer:
The ratio of s1:s2 is 1:9.
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@VIVEK9090
Explanation:
As particle initially at rest, initial velocity u is zero.
The body is in motion with uniform acceleration a.
AS WE KNOW,
[tex]s = ut + \frac{1}{2} a {t}^{2} [/tex]
For the displacement in 1st second, say s1
u= 0
t= 1
[tex]s1 = (0)(1) + \frac{1}{2} a ({1})^{2} [/tex]
That is
[tex]s1 = \frac{1}{2} a[/tex]
For the displacement in first 3 second, say s2
u=0
t= 3
[tex]s2 = (0)(3) + \frac{1}{2} a( {3})^{2} [/tex]
That is
[tex]s2 = \frac{9}{2} a[/tex]
Now take the ratio s1 : s2,
[tex] \frac{s1}{s2} = \frac{ \frac{1}{2}a }{ \frac{9}{2} a} = \frac{1}{9} [/tex]
Hence, the ratio of s1:s2 is 1:9.
Answer:
below
Explanation:
Let the acceleration of the particle be α.
Initial velocity v
i
=0
∴ Distance travelled in 2 seconds: x=v
i
t+
2
1
αt
2
where t=2 s
∴ x=0+
2
1
α(2)
2
=2α ..........(1)
Distance travelled in 4 seconds: S=v
i
t+
2
1
αt
2
where t=4 s
∴ S=0+
2
1
α(4)
2
=8α ............(2)
⟹ Distance traveled in last 2 seconds: y=8α−2α=6α
∴
x
y
=
2α
6α
⟹y=3x