Question

A body is moving in a straight line with initial velocity ‘u’ and uniform acceleration ‘a’. If the sum of the distances travelled in t and (t+1) seconds is 100cm, then its velocity after t seconds in cm/s is?

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Acceleration = a.
• Initial velocity = u.
• Distance travelled = 100 cm.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

Applying nth second formula,

$\large{\boxed{\tt S_n = u + \dfrac{a}{2}(2n - 1)}}$

Taking time = "t".

$\large{\tt S_t = u + \dfrac{a}{2}(2t - 1)}$

$\large{\tt S_t = u + \dfrac{a}{2}(2t - 1) \: ----(1)}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Applying nth second formula,

$\large{\boxed{\tt S_n = u + \dfrac{a}{2}(2n - 1)}}$

Taking time = "t + 1".

$\large{\tt S_{t +1}= u + \dfrac{a}{2}(2[t+1] - 1)}$

$\large{\tt S_{t + 1} = u + \dfrac{a}{2}(2t + 2 - 1)}$

$\large{\tt S_{t +1} = u + \dfrac{a}{2}(2t +1)}$

$\large{\tt S_{t+1}= u + \dfrac{a}{2}(2t + 1) \: ----(2)}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Adding equation (1) and equation (2).

$\large{\tt S_t + S_{t +1} = u + \dfrac{a}{2}(2t - 1) + u + \dfrac{a}{2}(2t + 1)}$

As we know,

$\large{\tt S_t + S_{t + 1} = 100}$

Substituting it,

$\large{\tt 100 = 2u + \dfrac{a}{2}(2t - 1 + 2t + 1)}$

$\large{\tt 100 = 2u + \dfrac{a}{2}(2t - \cancel{1} + 2t + \cancel{1})}$

It becomes,

$\large{\tt 100 = 2u + \dfrac{a}{2}(2t + 2t)}$

$\large{\tt 100 = 2u + \dfrac{a}{2}(4t)}$

$\large{\tt 100 = 2u + \dfrac{a}{\cancel{2}}(\cancel{4}t)}$

$\large{\tt 100 = 2u + a \times 2t}$

$\large{\tt 100 = 2u + 2at}$

$\large{\tt 2u + 2at = 100}$

$\large{\tt 2(u + at) = 100}$

$\large{\tt u + at = \dfrac{100}{2}}$

$\large{\tt u + at = \dfrac{\cancel{100}}{\cancel{2}}}$

$\large{\tt u + at = 50}$

Now,

$\large{\tt 50 = u + at}$

Comparing it with

$\large{\boxed{\tt v = u + at}}$

We get the velocity as,

$\huge{\boxed{\boxed{\tt v = 50 \: cm/s}}}$

So, the velocity after t seconds will be 50 cm/s.

$\rule{300}{1.5}$

Answer 2

Answer:

....Answer....

\huge{\bold{\underline{Given:-}}}

Given:−

Acceleration = a.

Initial velocity = u.

Distance travelled = 100 cm.

\huge{\bold{\underline{Explanation:-}}}

Explanation:−

\rule{300}{1.5}

Applying nth second formula,

\large{\boxed{\tt S_n = u + \dfrac{a}{2}(2n - 1)}}

S

n

=u+

2

a

(2n−1)

Taking time = "t".

\large{\tt S_t = u + \dfrac{a}{2}(2t - 1)}S

t

=u+

2

a

(2t−1)

\large{\tt S_t = u + \dfrac{a}{2}(2t - 1) \: ----(1)}S

t

=u+

2

a

(2t−1)−−−−(1)

\rule{300}{1.5}

\rule{300}{1.5}

Applying nth second formula,

\large{\boxed{\tt S_n = u + \dfrac{a}{2}(2n - 1)}}

S

n

=u+

2

a

(2n−1)

Taking time = "t + 1".

\large{\tt S_{t +1}= u + \dfrac{a}{2}(2[t+1] - 1)}S

t+1

=u+

2

a

(2[t+1]−1)

\large{\tt S_{t + 1} = u + \dfrac{a}{2}(2t + 2 - 1)}S

t+1

=u+

2

a

(2t+2−1)

\large{\tt S_{t +1} = u + \dfrac{a}{2}(2t +1)}S

t+1

=u+

2

a

(2t+1)

\large{\tt S_{t+1}= u + \dfrac{a}{2}(2t + 1) \: ----(2)}S

t+1

=u+

2

a

(2t+1)−−−−(2)

\rule{300}{1.5}

\rule{300}{1.5}

Adding equation (1) and equation (2).

\large{\tt S_t + S_{t +1} = u + \dfrac{a}{2}(2t - 1) + u + \dfrac{a}{2}(2t + 1)}S

t

+S

t+1

=u+

2

a

(2t−1)+u+

2

a

(2t+1)

As we know,

\large{\tt S_t + S_{t + 1} = 100}S

t

+S

t+1

=100

Substituting it,

\large{\tt 100 = 2u + \dfrac{a}{2}(2t - 1 + 2t + 1)}100=2u+

2

a

(2t−1+2t+1)

\large{\tt 100 = 2u + \dfrac{a}{2}(2t - \cancel{1} + 2t + \cancel{1})}100=2u+

2

a

(2t−

1

+2t+

1

)

It becomes,

\large{\tt 100 = 2u + \dfrac{a}{2}(2t + 2t)}100=2u+

2

a

(2t+2t)

\large{\tt 100 = 2u + \dfrac{a}{2}(4t)}100=2u+

2

a

(4t)

\large{\tt 100 = 2u + \dfrac{a}{\cancel{2}}(\cancel{4}t)}100=2u+

2

a

(

4

t)

\large{\tt 100 = 2u + a \times 2t}100=2u+a×2t

\large{\tt 100 = 2u + 2at}100=2u+2at

\large{\tt 2u + 2at = 100}2u+2at=100

\large{\tt 2(u + at) = 100}2(u+at)=100

\large{\tt u + at = \dfrac{100}{2}}u+at=

2

100

\large{\tt u + at = \dfrac{\cancel{100}}{\cancel{2}}}u+at=

2

100

\large{\tt u + at = 50}u+at=50

Now,

\large{\tt 50 = u + at}50=u+at

Comparing it with

\large{\boxed{\tt v = u + at}}

v=u+at

We get the velocity as,

\huge{\boxed{\boxed{\tt v = 50 \: cm/s}}}

v=50cm/s

So, the velocity after t seconds will be 50 cm/s.

$....Answer....</p><p></p><p>\huge{\bold{\underline{Given:-}}}Given:−</p><p></p><p>Acceleration = a.</p><p></p><p>Initial velocity = u.</p><p></p><p>Distance travelled = 100 cm.</p><p></p><p>\huge{\bold{\underline{Explanation:-}}}Explanation:−</p><p></p><p>\rule{300}{1.5}</p><p></p><p>Applying nth second formula,</p><p></p><p>\large{\boxed{\tt S_n = u + \dfrac{a}{2}(2n - 1)}}Sn=u+2a(2n−1)</p><p></p><p>Taking time = "t".</p><p></p><p>\large{\tt S_t = u + \dfrac{a}{2}(2t - 1)}St=u+2a(2t−1)</p><p></p><p>\large{\tt S_t = u + \dfrac{a}{2}(2t - 1) \: ----(1)}St=u+2a(2t−1)−−−−(1)</p><p></p><p>\rule{300}{1.5}</p><p></p><p>\rule{300}{1.5}</p><p></p><p>Applying nth second formula,</p><p></p><p>\large{\boxed{\tt S_n = u + \dfrac{a}{2}(2n - 1)}}Sn=u+2a(2n−1)</p><p></p><p>Taking time = "t + 1".</p><p></p><p>\large{\tt S_{t +1}= u + \dfrac{a}{2}(2[t+1] - 1)}St+1=u+2a(2[t+1]−1)</p><p></p><p>\large{\tt S_{t + 1} = u + \dfrac{a}{2}(2t + 2 - 1)}St+1=u+2a(2t+2−1)</p><p></p><p>\large{\tt S_{t +1} = u + \dfrac{a}{2}(2t +1)}St+1=u+2a(2t+1)</p><p></p><p>\large{\tt S_{t+1}= u + \dfrac{a}{2}(2t + 1) \: ----(2)}St+1=u+2a(2t+1)−−−−(2)</p><p></p><p>\rule{300}{1.5}</p><p></p><p>\rule{300}{1.5}</p><p></p><p>Adding equation (1) and equation (2).</p><p></p><p>\large{\tt S_t + S_{t +1} = u + \dfrac{a}{2}(2t - 1) + u + \dfrac{a}{2}(2t + 1)}St+St+1=u+2a(2t−1)+u+2a(2t+1)</p><p></p><p>As we know,</p><p></p><p>\large{\tt S_t + S_{t + 1} = 100}St+St+1=100</p><p></p><p>Substituting it,</p><p></p><p>\large{\tt 100 = 2u + \dfrac{a}{2}(2t - 1 + 2t + 1)}100=2u+2a(2t−1+2t+1)</p><p></p><p>\large{\tt 100 = 2u + \dfrac{a}{2}(2t - \cancel{1} + 2t + \cancel{1})}100=2u+2a(2t−1+2t+1)</p><p></p><p>It becomes,</p><p></p><p>\large{\tt 100 = 2u + \dfrac{a}{2}(2t + 2t)}100=2u+2a(2t+2t)</p><p></p><p>\large{\tt 100 = 2u + \dfrac{a}{2}(4t)}100=2u+2a(4t)</p><p></p><p>\large{\tt 100 = 2u + \dfrac{a}{\cancel{2}}(\cancel{4}t)}100=2u+2a(4t)</p><p></p><p>\large{\tt 100 = 2u + a \times 2t}100=2u+a×2t</p><p></p><p>\large{\tt 100 = 2u + 2at}100=2u+2at</p><p></p><p>\large{\tt 2u + 2at = 100}2u+2at=100</p><p></p><p>\large{\tt 2(u + at) = 100}2(u+at)=100</p><p></p><p>\large{\tt u + at = \dfrac{100}{2}}u+at=2100</p><p></p><p>\large{\tt u + at = \dfrac{\cancel{100}}{\cancel{2}}}u+at=2100</p><p></p><p>\large{\tt u + at = 50}u+at=50</p><p></p><p>Now,</p><p></p><p>\large{\tt 50 = u + at}50=u+at</p><p></p><p>Comparing it with</p><p></p><p>\large{\boxed{\tt v = u + at}}v=u+at</p><p></p><p>We get the velocity as,</p><p></p><p>\huge{\boxed{\boxed{\tt v = 50 \: cm/s}}}v=50cm/s</p><p></p><p>So, the velocity after t seconds will be 50 cm/s.</p><p></p><p>$