Question

Qus. An object starting from rest travels 20 meter in first 2sec And 160 meter in next 4 sec. What will be the velocity after 7sec from the start point.

Answer 1

Answer:

if it (an object) travels

20m.in first 2sec.

160m.in next 4 sec.means its velocity is 160m./sec

after 6sec.it get started.

then

160m.=in 6th sec.

after 7th second

speed =x

then

160m.=in 6th sec.

xm.=in 7th sec.

7×160÷6

=186.62m/sec.

therefore the velocity of that object on 7th sec. from the starting point is 186.62m./sec

Answer 2

Given:

Initial speed of object, u= 0 m/s

(Since it starts from rest)

Distance covered by object in first 2s, S₁= 20m

Distance covered by object in next 4s, S₂= 160m

To Find:

Velocity of given object after 7s from starting point

Solution:

We know that,

• According to first equation of motion for constant acceleration

$\purple{\boxed{\bf{v=u+at}}}$

• According to second equation of motion for constant acceleration

$\pink{\boxed{\bf{s=ut+\frac{1}{2}at^{2}}}}$

where,

u is the initial velocity of body

v is the final velocity of body

a is acceleration of body

t is time taken by body

s is displacement of body in given time

$\rule{190}{2}$

Now,

Let the acceleration of body for first 2s be a₁ and acceleration of body for next 4s be a₂

On applying second equation of motion on given body for the first 2s, we get

$\longrightarrow\rm{S_{1}=0(2)+\dfrac{1}{2}a_{1}(2)^{2}}$

$\longrightarrow\rm{20=\dfrac{1}{\cancel{2}}a_{1}(\cancel{4})}$

$\longrightarrow\rm{20=2a_{1}}$

$\longrightarrow\rm{2a_{1}=20}$

$\longrightarrow\rm{a_{1}=\dfrac{20}{2}}$

$\longrightarrow\rm{a_{1}=10\:m/s^{2}}$

Let the velocity of object after 2s be v'

On applying first equation of motion on given body for the first 2s, we get

$\longrightarrow\rm{v'=0+10(2)}$

$\longrightarrow\rm{v'=20\:m/s}$

$\rule{190}{2}$

Now, This v' is the initial velocity of next 4s

So, on applying second equation of motion on given body for the next 4s, we get

$\longrightarrow\rm{S_{2}=20(4)+\dfrac{1}{2}a_{2}(4)^{2}}$

$\longrightarrow\rm{160=80+\dfrac{1}{\cancel{2}}a_{2}(\cancel{16})}$

$\longrightarrow\rm{160=80+8a_{2}}$

$\longrightarrow\rm{160-80=8a_{2}}$

$\longrightarrow\rm{80=8a_{2}}$

$\longrightarrow\rm{8a_{2}=80}$

$\longrightarrow\rm{a_{2}=\dfrac{80}{8}}$

$\longrightarrow\rm{a_{2}=10\:m/s}$

$\rule{190}{2}$

Since, a₁=a₂

This means given object is having constant acceleration of 10 m/s² throughout the journey

∴ we can now apply first equation of motion for whole journey

Let the velocity of given object after 7s from the starting point be v

So, on using first equation of motion for whole journey, we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{v=0+(10)7}$

$\longrightarrow\rm\green{v=70\:m/s}$

Hence, the velocity of given object after 7s from the starting point is 70 m/s .