Question

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:
$V_{avg} = u+v/2$
$s = [\frac{u+v}{2} ]t$
$s = ut+\frac{1}{2}at^{2}$
$s = vt-\frac{1}{2}at^{2}$
$t = \frac{v-u}{a}$
$v^{2} -u^{2} = 2as$

s = displacement
a = acceleration
u = initial velocity
v = final velocity
t = time

Answer 1

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,

$s = ut + \frac{1}{2} a {t}^{2}$

For the displacement in 1st second, say s1

u= 0

t= 1

$s1 = (0)(1) + \frac{1}{2} a ({1})^{2}$

That is

$s1 = \frac{1}{2} a$

For the displacement in first 3 second, say s2

u=0

t= 3

$s2 = (0)(3) + \frac{1}{2} a( {3})^{2}$

That is

$s2 = \frac{9}{2} a$

Now take the ratio s1 : s2,

$\frac{s1}{s2} = \frac{ \frac{1}{2}a }{ \frac{9}{2} a} = \frac{1}{9}$

Hence, the ratio of s1:s2 is 1:9.

Answer 2

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