A train Start from one station reaches second station in 3 hr which is at 48 km away from first station. And then returns back to the first station and takes 5 hr. Find the Average speed and average velocity of the train?
Answers
Explanation:
As per the provided information in the given question, we have :
Time taken by train from one station (Station 1) to second station (Station 2) = 3 hrs
Time taken by train from second station (Station 2) to one station (Station 1) = 5 hrs
Distance between station one and second station = 48 km
We are asked to calculate average speed and average velocity of the train.
★ Calculating average speed of the train :
Average speed refers to the total distance covered divided by total time.
begin gathered longrightarrow quad \pmb{\boxed{\sf {Speed_{(avg)} = \dfrac{Total \; distance}{Total \; time }}}}\\ \end{gathered}
⟶
Speed
(avg)
=
Totaltime
Totaldistance
Speed
(avg)
=
Totaltime
Totaldistance
\begin{gathered} \\ \longrightarrow \sf{\quad Speed_{(avg)} = \dfrac{Distance_{(Station \; 1 \; to \; 2)} + Distance_{(Station \; 2 \; to \; 1) }}{Time_{(Station \; 1 \; to \; 2)} +Time_{(Station \; 2 \; to \; 1)}} } \\ \end{gathered}
⟶Speed
(avg)
=
Time
(Station1to2)
+Time
(Station2to1)
Distance
(Station1to2)
+Distance
(Station2to1)
\begin{gathered} \\ \longrightarrow \sf{\quad {Speed_{(avg)} = \dfrac{(48 + 48) \; km }{(3 + 5) \; hrs} }} \\ \end{gathered}
⟶Speed
(avg)
=
(3+5)hrs
(48+48)km
\begin{gathered} \\ \longrightarrow \sf{\quad {Speed_{(avg)} =\cancel{ \dfrac{ 96 \; km }{ 8\; hrs}} }} \\ \end{gathered}
⟶Speed
(avg)
=
8hrs
96km
\begin{gathered} \\ \longrightarrow \bf{\quad {\underline{ Speed_{(avg)} = 12 \; km/h}}} \\ \end{gathered}
⟶
Speed
(avg)
=12km/h
Therefore, average speed of the train is 12 km/h.
★ Calculating average velocity :
Average velocity refers to the total displacement divided by total time.
\begin{gathered} \\ \longrightarrow \quad \pmb{\boxed{\sf {Velocity_{(avg)} = \dfrac{Total \; displacement}{Total \; time }}}}\\ \end{gathered}
⟶
Velocity
(avg)
=
Totaltime
Totaldisplacement
Velocity
(avg)
=
Totaltime
Totaldisplacement
The train came back to its initial position after covering certain distance. Whenever tha body comes back to its initial position after covering certain distance, then its displacement is 0.
\begin{gathered} \\ \longrightarrow \sf{\quad {Velocity_{(avg)} = \dfrac{0\; km }{ (3 + 5) \; hrs} }} \\ \end{gathered}
⟶Velocity
(avg)
=
(3+5)hrs
0km
\begin{gathered} \\ \longrightarrow \sf{\quad {Velocity_{(avg)} =\cancel{ \dfrac{ 0 \; km }{ 8\; hrs}} }} \\ \end{gathered}
⟶Velocity
(avg)
=
8hrs
0km
\begin{gathered} \\ \longrightarrow \bf{\quad {\underline{ Velocity_{(avg)} = 0 \; km/h}}} \\ \end{gathered}
⟶
Velocity
(avg)
=0km/h
Therefore, average velocity of the train is 0 km/h.