Question Statement

An electric train starts from rest at station $A$ and moves along a straight level track. The motion consists of three stages:

Acceleration: The train accelerates uniformly at $0.4 m/s^{2}$ from rest to a speed of $16 m/s$
Constant Velocity: The speed of $16 m/s$ is maintained for a displacement of $2000 m$
Deceleration: The train retards uniformly for $20 s$ before coming to rest at station $B$

Find: (i) The total time taken for the journey from $A$ to $B$ (ii) The total displacement from $A$ to $B$ (iii) Sketch the displacement-time graph, showing clearly the shape of the graph for each stage of the journey

Background and Explanation

This problem involves analyzing motion with constant acceleration (and deceleration) using the equations of kinematics. The journey must be broken into three distinct phases, where the acceleration changes between stages, requiring separate calculations for time and displacement in each segment.

Solution

Part (i): Total Time Taken

Stage 1: Acceleration Phase Using the definition of acceleration: $a = \frac{Δ v}{t}$

Substituting $a = 0.4 m/s^{2}$ and $Δ v = 16 - 0 = 16 m/s$ : $0.4 = \frac{16}{t _{1}}$

Solving for $t_{1}$ : $t_{1} = \frac{16}{0.4} = 40 s$

Stage 2: Constant Velocity Phase Using $s = v \times t$ with $s = 2000 m$ and $v = 16 m/s$ : $2000 = 16 \times t_{2}$

Solving for $t_{2}$ : $t_{2} = \frac{2000}{16} = 125 s$

Stage 3: Deceleration Phase The time is given directly: $t_{3} = 20 s$

Total Time: $Total time = t_{1} + t_{2} + t_{3} = 40 + 125 + 20 = 185 s$

Part (ii): Total Displacement from $A$ to $B$

Displacement during Stage 1 (Acceleration): Using the equation $v^{2} = u^{2} + 2 a s$ , rearranged as $s = \frac{v ^{2} - u ^{2}}{2 a}$ :

$s_{1} = \frac{1 6 ^{2} - 0 ^{2}}{2 \times 0.4} = \frac{256}{0.8} = 320 m$

Displacement during Stage 2 (Constant Velocity): Given directly in the problem: $s_{2} = 2000 m$

Displacement during Stage 3 (Deceleration): First, find the deceleration using $a = \frac{v _{f} - v _{i}}{t}$ : $a = \frac{0 - 16}{20} = - 0.8 m/s^{2}$

Now apply $s = \frac{v _{f}^{2} - v _{i}^{2}}{2 a}$ : $s_{3} = \frac{0 ^{2} - 1 6 ^{2}}{2 \times ( - 0.8 )} = \frac{- 256}{- 1.6} = 160 m$

Total Displacement: $Total displacement = s_{1} + s_{2} + s_{3} = 320 + 2000 + 160 = 2480 m$

Part (iii): Displacement-Time Graph

The displacement-time graph consists of three distinct sections:

First section (0 to 40 s): Curved (parabolic) with increasing gradient, starting from origin (0,0) and reaching (40, 320). The curve is concave down since acceleration is positive but decreasing? Wait, actually for constant positive acceleration from rest, $s \propto t^{2}$ , so it's a parabola opening upward (increasing slope).
Second section (40 s to 165 s): Straight line with constant positive gradient from (40, 320) to (165, 2320), representing constant velocity.
Third section (165 s to 185 s): Curved (parabolic) with decreasing gradient, ending at (185, 2480) with zero gradient (horizontal), representing deceleration to rest.

Key Formulas or Methods Used

$a = \frac{Δ v}{Δ t}$ (definition of acceleration)
$v = u + a t$ (velocity-time relation)
$s = \frac{v ^{2} - u ^{2}}{2 a}$ (velocity-displacement relation for constant acceleration)
$s = v t$ (distance at constant velocity)
Analysis of motion by dividing journey into distinct phases with different acceleration values

Summary of Steps

Calculate time for acceleration phase using $t = \frac{v - u}{a}$ with $u = 0$ , $v = 16$ , and $a = 0.4$
Calculate time for constant velocity phase using $t = \frac{s}{v}$ with $s = 2000$ and $v = 16$
Sum all three times (acceleration + constant velocity + given deceleration time of 20s) to get total time
Calculate displacement for acceleration phase using $s = \frac{v ^{2} - u ^{2}}{2 a}$
Record given displacement for constant velocity phase (2000 m)
Calculate deceleration using $a = \frac{v _{f} - v _{i}}{t}$ with $v_{f} = 0$ , $v_{i} = 16$ , and $t = 20$
Calculate displacement for deceleration phase using $s = \frac{v _{f}^{2} - v _{i}^{2}}{2 a}$ with the calculated deceleration
Sum all three displacements to get total displacement from A to B
Sketch the s-t graph showing: parabolic curve (increasing slope) for acceleration, straight line for constant velocity, and parabolic curve (decreasing slope to zero) for deceleration

Question Statement

An electric train starts from rest at station $A$ and moves along a straight level track. The motion consists of three stages:

Acceleration: The train accelerates uniformly at $0.4 m/s^{2}$ from rest to a speed of $16 m/s$
Constant Velocity: The speed of $16 m/s$ is maintained for a displacement of $2000 m$
Deceleration: The train retards uniformly for $20 s$ before coming to rest at station $B$

Background and Explanation

Solution

Part (i): Total Time Taken

Stage 1: Acceleration Phase Using the definition of acceleration: $a = \frac{Δ v}{t}$

Substituting $a = 0.4 m/s^{2}$ and $Δ v = 16 - 0 = 16 m/s$ : $0.4 = \frac{16}{t _{1}}$

Solving for $t_{1}$ : $t_{1} = \frac{16}{0.4} = 40 s$

Stage 2: Constant Velocity Phase Using $s = v \times t$ with $s = 2000 m$ and $v = 16 m/s$ : $2000 = 16 \times t_{2}$

Solving for $t_{2}$ : $t_{2} = \frac{2000}{16} = 125 s$

Stage 3: Deceleration Phase The time is given directly: $t_{3} = 20 s$

Total Time: $Total time = t_{1} + t_{2} + t_{3} = 40 + 125 + 20 = 185 s$

Part (ii): Total Displacement from $A$ to $B$

Displacement during Stage 1 (Acceleration): Using the equation $v^{2} = u^{2} + 2 a s$ , rearranged as $s = \frac{v ^{2} - u ^{2}}{2 a}$ :

$s_{1} = \frac{1 6 ^{2} - 0 ^{2}}{2 \times 0.4} = \frac{256}{0.8} = 320 m$

Displacement during Stage 2 (Constant Velocity): Given directly in the problem: $s_{2} = 2000 m$

Displacement during Stage 3 (Deceleration): First, find the deceleration using $a = \frac{v _{f} - v _{i}}{t}$ : $a = \frac{0 - 16}{20} = - 0.8 m/s^{2}$

Now apply $s = \frac{v _{f}^{2} - v _{i}^{2}}{2 a}$ : $s_{3} = \frac{0 ^{2} - 1 6 ^{2}}{2 \times ( - 0.8 )} = \frac{- 256}{- 1.6} = 160 m$

Total Displacement: $Total displacement = s_{1} + s_{2} + s_{3} = 320 + 2000 + 160 = 2480 m$

Part (iii): Displacement-Time Graph

The displacement-time graph consists of three distinct sections:

First section (0 to 40 s): Curved (parabolic) with increasing gradient, starting from origin (0,0) and reaching (40, 320). The curve is concave down since acceleration is positive but decreasing? Wait, actually for constant positive acceleration from rest, $s \propto t^{2}$ , so it's a parabola opening upward (increasing slope).
Second section (40 s to 165 s): Straight line with constant positive gradient from (40, 320) to (165, 2320), representing constant velocity.
Third section (165 s to 185 s): Curved (parabolic) with decreasing gradient, ending at (185, 2480) with zero gradient (horizontal), representing deceleration to rest.

Key Formulas or Methods Used

$a = \frac{Δ v}{Δ t}$ (definition of acceleration)
$v = u + a t$ (velocity-time relation)
$s = \frac{v ^{2} - u ^{2}}{2 a}$ (velocity-displacement relation for constant acceleration)
$s = v t$ (distance at constant velocity)
Analysis of motion by dividing journey into distinct phases with different acceleration values

Summary of Steps

Calculate time for acceleration phase using $t = \frac{v - u}{a}$ with $u = 0$ , $v = 16$ , and $a = 0.4$
Calculate time for constant velocity phase using $t = \frac{s}{v}$ with $s = 2000$ and $v = 16$
Sum all three times (acceleration + constant velocity + given deceleration time of 20s) to get total time
Calculate displacement for acceleration phase using $s = \frac{v ^{2} - u ^{2}}{2 a}$
Record given displacement for constant velocity phase (2000 m)
Calculate deceleration using $a = \frac{v _{f} - v _{i}}{t}$ with $v_{f} = 0$ , $v_{i} = 16$ , and $t = 20$
Calculate displacement for deceleration phase using $s = \frac{v _{f}^{2} - v _{i}^{2}}{2 a}$ with the calculated deceleration
Sum all three displacements to get total displacement from A to B
Sketch the s-t graph showing: parabolic curve (increasing slope) for acceleration, straight line for constant velocity, and parabolic curve (decreasing slope to zero) for deceleration

5.1 Q-4

Question Statement

Background and Explanation

Solution

Part (i): Total Time Taken

Part (ii): Total Displacement from $A$ to $B$

Part (iii): Displacement-Time Graph

Key Formulas or Methods Used

Summary of Steps

5.1 Q-4

Question Statement

Background and Explanation

Solution

Part (i): Total Time Taken

Part (ii): Total Displacement from $A$ to $B$

Part (iii): Displacement-Time Graph

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Part (i): Total Time Taken

Part (ii): Total Displacement from A to B

Part (iii): Displacement-Time Graph

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Part (i): Total Time Taken

Part (ii): Total Displacement from A to B

Part (iii): Displacement-Time Graph

Key Formulas or Methods Used

Summary of Steps

Part (ii): Total Displacement from $A$ to $B$

Part (ii): Total Displacement from $A$ to $B$