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Answers
Answer
- 2.5 meters
Question
A particle moving with uniform acceleration in a straight line covers 3 m in 8th second and 5 m in the 16th second of it's motion . Find the distance travelled by it from the beginning of the 6th second to the end of it's 15th second
Solution
The distance travelled by a body during the nth second is given by ,
In 8th second , it covers 3 m ,
In 16th second , it covers 5 m ,
Solve (2) - (1) ,
On sub. a value in (2) , we get ,
Now , Distance travelled by the given particle from the beginning of the 6th second to the end of it's 15th second is given by ,
Answer:
Answer
2.5 meters
Question
A particle moving with uniform acceleration in a straight line covers 3 m in 8th second and 5 m in the 16th second of it's motion . Find the distance travelled by it from the beginning of the 6th second to the end of it's 15th second
Solution
The distance travelled by a body during the nth second is given by ,
\bigstar\ \; \bf S_n=u+\dfrac{a}{2}(2n-1)\ \; \bigstar★ S
n
=u+
2
a
(2n−1) ★
In 8th second , it covers 3 m ,
\begin{gathered}\to \rm S_8=u+\dfrac{a}{2}(2(8)-1)\\\\\to \rm 3=u+\dfrac{a}{2}(15)\\\\\to \rm 3=u+\dfrac{15a}{2}...(1)\end{gathered}
→S
8
=u+
2
a
(2(8)−1)
→3=u+
2
a
(15)
→3=u+
2
15a
...(1)
In 16th second , it covers 5 m ,
\begin{gathered}\to \rm S_{16}=u+\dfrac{a}{2}(2(16)-1)\\\\\to \rm 5=u+\dfrac{a}{2}(31)\\\\\to \rm 5=u+\dfrac{31a}{2}...(2)\end{gathered}
→S
16
=u+
2
a
(2(16)−1)
→5=u+
2
a
(31)
→5=u+
2
31a
...(2)
Solve (2) - (1) ,
\begin{gathered}\to \rm 2=\dfrac{16a}{2}\\\\\to \rm 8a=2\\\\\to \rm a=\dfrac{1}{4}\ m/s^2\end{gathered}
→2=
2
16a
→8a=2
→a=
4
1
m/s
2
On sub. a value in (2) , we get ,
\begin{gathered}\to \rm 5=u+\dfrac{31(\frac{1}{4})}{2}\\\\\to \rm 5=u+\dfrac{31}{8}\\\\\to \rm u=\dfrac{9}{8}\ m/s\end{gathered}
→5=u+
2
31(
4
1
)
→5=u+
8
31
→u=
8
9
m/s
Now , Distance travelled by the given particle from the beginning of the 6th second to the end of it's 15th second is given by ,
\begin{gathered}\to \rm S_{16}-S_{6}\\\\\to \rm \left(\dfrac{9}{8}+\dfrac{\frac{1}{4}}{2}(2(16)-1) \right)-\left( \dfrac{9}{8}+\dfrac{\frac{1}{4}}{2}(2(6)-1) \right)\\\\\to \rm \dfrac{9}{8}+\dfrac{31}{8}-\dfrac{9}{8}-\dfrac{11}{8}\\\\\to \rm \dfrac{20}{8}\\\\\to \rm 2.5\ m\ \bigstar\end{gathered}
→S
16
−S
6
→(
8
9
+
2
4
1
(2(16)−1))−(
8
9
+
2
4
1
(2(6)−1))
→
8
9
+
8
31
−
8
9
−
8
11
→
8
20
→2.5 m ★