व्हिच ऑफ द फॉलोइंग कलेक्शन इन सेट गुड बॉयज इन योर क्लास डिफिकल्ट एग्जांपल्स इन योर टेक्सबुक प्राइम नंबर्स बिटवीन 1200 इंटेलीजेंट पीपल इन योर साइट
Answers
Answer:
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\blue{\bold{\underline{\underline{Answer:}}}}
Answer:
\green{\tt{\therefore{Distance\:travelled=625\:m}}}∴Distancetravelled=625m
\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}
Step−by−stepexplanation:
\begin{gathered}\green{\underline \bold{Given :}} \\ \tt: \implies Initial \: speed(u) = 90 \: km/h \\ \\ \tt: \implies Acceleration(a) = - 0.5 \: {m/s}^{2} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Distance \: travel(s) = ?\end{gathered}
Given:
:⟹Initialspeed(u)=90km/h
:⟹Acceleration(a)=−0.5m/s
2
ToFind:
:⟹Distancetravel(s)=?
• According to given question :
\begin{gathered}\tt \circ \: Initial \: speed = 90 \times \frac{5}{18} = 25 \: m/s \\ \\ \tt \circ \: Final \: speed = 0 \: m/s\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {25}^{2} + 2 \times - 0.5 \times s \\ \\ \tt: \implies 0 = 625 - s \\ \\ \tt: \implies - 625 = - s \\ \\ \green{\tt: \implies s = 625 \: m} \\ \\ \green{\tt \therefore Distance \: travelled \: by \: train \: is \: 625 \: m \: after \: applying \: brake}\end{gathered}
∘Initialspeed=90×
18
5
=25m/s
∘Finalspeed=0m/s
Asweknowthat
:⟹v
2
=u
2
+2as
:⟹0
2
=25
2
+2×−0.5×s
:⟹0=625−s
:⟹−625=−s
:⟹s=625m
∴Distancetravelledbytrainis625mafterapplyingbrake
\begin{gathered}\blue{\huge {\boxed{ \tt Some \: related \: formula}}} \\ \\ \orange{ \tt \circ \: First \: eqn \: of \: motion \to \: v = u + at} \\ \\ \orange{ \tt \circ \: Second\: eqn \: of \: motion \to \: s= u t+ \frac{1}{2} {at}^{2} } \\ \\ \orange{ \tt \circ \: Time \: of \: flight \to \: t= \frac{ 2u \: {sin} \: \theta }{g} } \\ \\ \orange{ \tt \circ \: Range \to \: R = \frac{ {u}^{2} sin \: 2 \theta}{g} } \\ \\ \orange{ \tt \circ \: Maximum \: height \to \: H_{max}= \frac{ {u}^{2} {sin}^{2} \: \theta}{2g} }\end{gathered}
Somerelatedformula
∘Firsteqnofmotion→v=u+at
∘Secondeqnofmotion→s=ut+
2
1
at
2
∘Timeofflight→t=
g
2usinθ
∘Range→R=
g
u
2
sin2θ
∘Maximumheight→H
max
=
2g
u
2
sin
2
θ