A train 120m long moving on a straight and level track with uniform speed passes a pole in 6 seconds. Find the time it will take to cross a 50m long bridge.
Answers
Answer:
As, The train is 120 m long and bridge is 50 m long = {120+ 50 = 170 m } So, Time taken = distance/speed
= 170/20
= 8.5 seconds
Mark me the Brainiest if you liked my answer.
Answer:
pls forgive me I wrote wrongly your last question answer is
Distance = 27 km
Displacement = 13 km
Explanation:
As per the provided information in the given question, we have (refer to the attachment for better understanding) :
An object goes 10km towards East (AB).
Then, 5 km towards west. (BC)
After that, it goes 12 km towards north. (CD)
★ Calculating distance travelled :
Distance means total length of the path covered by the body in the entire journey. So,
\begin{gathered} \\ \longrightarrow \quad \sf {Distance = AB + BC + CD } \\\end{gathered}
⟶Distance=AB+BC+CD
\begin{gathered} \\ \longrightarrow \quad \sf {Distance = (10 + 5 + 12) \; km } \\\end{gathered}
⟶Distance=(10+5+12)km
\begin{gathered} \\ \longrightarrow \quad \sf {Distance = (15 + 12) \; km } \\\end{gathered}
⟶Distance=(15+12)km
\begin{gathered} \\ \longrightarrow \quad \bf \underline {Distance = 27 \; km } \\\end{gathered}
⟶
Distance=27km
Therefore, distance travelled is 27 km.
★ Calculating displacement :
Displacement is the shortest distance from initial position to final position.
Here, A is its initial position and D is its final position. Shortest distance from A to D is a straight line that is AD. We'll be using Pythagoras theorem to find the length of ADm Before that, we need to find the length of AC.
\begin{gathered} \\ \longrightarrow \quad \sf {AC = AB - BC } \\\end{gathered}
⟶AC=AB−BC
\begin{gathered} \\ \longrightarrow \quad \sf {AC = (10 - 5) \; km } \\\end{gathered}
⟶AC=(10−5)km
\begin{gathered} \\ \longrightarrow \quad \bf \underline {AC = 5 \; km } \\\end{gathered}
⟶
AC=5km
Now, by using Pythagoras theorem,
\begin{gathered} \\ \longrightarrow \quad \bf {AD^2 = AC^2 + CD^2 } \\\end{gathered}
⟶AD
2
=AC
2
+CD
2
\begin{gathered} \\ \longrightarrow \quad \sf {(Displacement)^2 = (5 \; km)^2 + (12 \; km)^2 } \\\end{gathered}
⟶(Displacement)
2
=(5km)
2
+(12km)
2
\begin{gathered} \\ \longrightarrow \quad \sf {(Displacement)^2 = 25 \; km^2 + 144 \; km^2 } \\\end{gathered}
⟶(Displacement)
2
=25km
2
+144km
2
\begin{gathered} \\ \longrightarrow \quad \sf {(Displacement)^2 = 169 \; km^2 } \\\end{gathered}
⟶(Displacement)
2
=169km
2
\begin{gathered} \\ \longrightarrow \quad \sf {Displacement = \sqrt{169 \; km^2 } } \\\end{gathered}
⟶Displacement=
169km
2
\begin{gathered} \\ \longrightarrow \quad \bf \underline {Displacement = 13 \; km } \\\end{gathered}
⟶
Displacement=13km
Therefore, the displacement is 13 km.