the velocity time graph of a train is depicted in the figure. the avg velocity in time od
Answers
Answer:
Solution
\Large{\underline{\mathfrak{\bf{Given}}}}
Given
Distance of object from concave mirror (U) = -10 cm
Radius of curvature of concave mirror ( R ) = -8 cm = 2F
So, F = -8/2 = -4 cm
\Large{\underline{\mathfrak{\bf{Find}}}}
Find
Distance of image (V) .
\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}
Explanation
\Large{\underline{\mathfrak{\bf{Using\:mirror\:formula}}}}
Usingmirrorformula
\red{\boxed{\sf{\orange{\:\dfrac{1}{F}\:=\:\dfrac{1}{U}\:+\:\dfrac{1}{V}}}}}
F
1
=
U
1
+
V
1
Substitute value of U and F in above equation ,
\begin{gathered}:\implies\sf{\:\dfrac{1}{V}\:+\:\dfrac{1}{-10}\:=\:\dfrac{1}{-4}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:-\:\dfrac{1}{4}\:+\:\dfrac{1}{10}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:\dfrac{-10+4}{40}} \\ \\ :\implies\sf{\:\dfrac{1}{V}\:=\:\dfrac{-14}{40}} \\ \\ :\implies\sf{\:V\:=\:\dfrac{-40}{14}} \\ \\ :\implies\sf{\red{\:V\:=\:\dfrac{-20}{7}\:cm}}\end{gathered}
:⟹
V
1
+
−10
1
=
−4
1
:⟹
V
1
=−
4
1
+
10
1
:⟹
V
1
=
40
−10+4
:⟹
V
1
=
40
−14
:⟹V=
14
−40
:⟹V=
7
−20
cm
\Large{\underline{\mathfrak{\bf{Thus}}}}
Thus
Distance of image (V ) = -20/7 cm
When, the image distance is negative , the image is behind the mirror .
So, the image is virtual and upright.
Answer:
Total distance travelled = Area of trapezium OABC =
2
1
×(OC+AB)×AD=
2
1
×(35+20)×1000=55×500=27500m=27.5km
∴ average velocity =
totaltime
totaldistance
=
35
27500
=785.7m.min
−1
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