Question

A train starting from rest, picks up a velocity of 20 m/s in 200 s. It continues to move at he same rate for the next 500 s, and is then brought to rest in another 100 s.
(i) plot a velocity-time graph. (ii) from the graph, calculate : (a) the uniform rate of
acceleration, (b) the uniform rate of retardation, (c) the total distance covered before stopping, (d)the average velocity over the total journey.

Answer 1

A train starting from rest, picks up a velocity of 20 m/s in 200 s. It continues to move at he same rate for the next 500 s, and is then brought to rest in another 100 s.

(i) plot a velocity-time graph.

Graph is in the attachment

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(ii) from the graph, calculate :

(a) the uniform rate of acceleration

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{Final \: velocity - Initial \: velocity }{Time \: interval} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{f- i }{t} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{20 - 0}{200} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{20 }{200} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{2 }{20} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{ \cancel{2 }}{ \cancel{20}} = \dfrac{1}{10}m {s}^{ - 2} \end{gathered} \end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \blue{0.1 }\end{gathered} \end{gathered} \end{gathered}$

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(b) the uniform rate of retardation

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{ Initial \: velocity - Final \: velocity - }{Time \: interval} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{i - f}{t} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{0 - 20}{100} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{ \cancel{ - 20}}{ \cancel{100}} = \dfrac{ - 1 }{5} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \purple{ - 0.2 {ms}^{ - 2} }\end{gathered} \end{gathered} \end{gathered}$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀

(c) the total distance covered before stopping.

as you can see in the attachment that we have divided it into three parts .

Distance covered - I part + II part + III part

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \left(\dfrac{1}{2} \times 200 \times 20 \right)+ \left( 500 \times 20 \right)+ \left(\frac{1}{2} \times 100 \times 20 \right)\end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \left( 2000 \right)+ \left( 10000 \right)+ \left( 1000 \right)\end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \pink{ 13000 {m} }\end{gathered} \end{gathered} \end{gathered}$

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(d)the average velocity over the total journey.

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{Total \: distance }{ Total \: time} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{13000 }{ 800} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \dfrac{130 \cancel{00} }{ 8 \cancel{00}} \end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered} \begin{gathered}\\ \sf \mapsto \red{ 16.25m}\end{gathered} \end{gathered} \end{gathered}$

Answer 2

above ans is correct .

hi friend nice to meet you