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?
Answers
- Acceleration = a.
- Initial velocity = u.
- Distance travelled = 100 cm.
Applying nth second formula,
Taking time = "t".
Applying nth second formula,
Taking time = "t + 1".
Adding equation (1) and equation (2).
As we know,
Substituting it,
It becomes,
Now,
Comparing it with
We get the velocity as,
So, the velocity after t seconds will be 50 cm/s.
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.