Question

A train starts from rest and accelerates uniformly at a rate of $2m{{s}^{-2}}$ for $10\;s$ . It then maintains a constant speed for $200\;s$ . The brakes are then applied and the train is uniformly retarded and comes to rest in $50\;s$ . Find the total distance travelled.
(A) $4600\;km$
(B) $460\;m$
(C) $46\;m$
(D) $4600\;m$

Answer 1

Here, the train accelerates uniformly. Hence we can use the equations for the uniformly accelerated motion for a straight line motion to find the velocity and distance travelled in all three cases.

Complete answer:
Let us note the given data;
As the train starts from rest, initial velocity for first case ${{u}_{1}}=0m{{s}^{-1}}$
Uniform acceleration for the first case ${{a}_{1}}=2m{{s}^{-2}}$
Final velocity achieved at the end of acceleration for the first case ${{v}_{1}}=?$
Time period during which the train accelerates for the first case ${{t}_{1}}=10s$
Distance travelled by train with uniform acceleration during the first case ${{s}_{1}}=?$
Initial velocity for the second case which is final velocity for first case ${{u}_{2}}={{v}_{1}}=?$
As train maintains uniform speed, acceleration for the second case ${{a}_{2}}=0m{{s}^{-2}}$
As the speed is uniform, final velocity for the second case which is equal to the initial velocity for the second case ${{v}_{2}}={{u}_{2}}=?$
Time period during which the speed remains constant for the second case ${{t}_{2}}=200s$
Distance travelled by train with uniform speed in the second case ${{s}_{2}}=?$
Initial velocity for the third case which is final velocity for second case ${{u}_{3}}={{v}_{2}}=?$
As brakes are applied, the uniform deceleration for the third case ${{a}_{3}}=?$
As the train finally stops, final velocity for the third case ${{v}_{3}}=0m{{s}^{-1}}$
Time period for which train decelerates for the third case ${{t}_{3}}=50s$
Distance travelled by train with uniform deceleration for the third case ${{s}_{3}}=?$
The equations for uniformly accelerated motion in a straight line are defined as,
Equation to be used if we don’t require or don’t have the value of distance travelled is $v=u+at$
Equation to be used if we don’t require or don’t have the value of final velocity is $s=ut+\dfrac{1}{2}a{{t}^{2}}$
Equation to be used if we don’t require or don’t have the value of time taken ${{v}^{2}}={{u}^{2}}+2as$
Equation to be used if we don’t require or don’t have the value of acceleration $s=\left( \dfrac{u+v}{2} \right)t$
The last equation is not predefined but can be derived from the first three equations.
Now, for the first case we have value of initial velocity, acceleration, and time taken.
By using equation $(I)$ we can find the final velocity as,
${{v}_{1}}={{u}_{1}}+{{a}_{1}}{{t}_{1}}$
Substituting the given values,
$\therefore {{v}_{1}}=0+(2)(10)$
$\therefore {{v}_{1}}=20m{{s}^{-1}}$
By using equation $(II)$ we can find the distance travelled as,
${{s}_{1}}={{u}_{1}}{{t}_{1}}+\dfrac{1}{2}{{a}_{1}}{{t}_{1}}^{2}$
Substituting the given values,
$\therefore {{s}_{1}}=(0)(10)+\dfrac{1}{2}(2){{(10)}^{2}}$
$\therefore {{s}_{1}}=100m$
For the second case, we have initial velocity, final velocity, acceleration, and time taken.
By using the equation $(II)$ we can find the distance travelled as,
${{s}_{2}}={{u}_{2}}{{t}_{2}}+\dfrac{1}{2}{{a}_{2}}{{t}_{2}}^{2}$
$\therefore {{s}_{2}}=(20)(200)+\dfrac{1}{2}(0){{(200)}^{2}}$
$\therefore {{s}_{2}}=4000m$
For the third case we have the final velocity, initial velocity, and the time taken.
By using the equation $(IV)$ we can find the distance travelled as,
${{s}_{3}}=\left( \dfrac{{{u}_{3}}+{{v}_{3}}}{2} \right){{t}_{3}}$
Substituting the given values,
$\therefore {{s}_{3}}=\left( \dfrac{20+0}{2} \right)(50)$
$\therefore {{s}_{3}}=500m$
Now, the total distance travelled is
$\therefore s={{s}_{1}}+{{s}_{2}}+{{s}_{3}}$
Substituting the obtained values,
$\therefore s=100m+4000m+500m$
$\therefore s=4600m$
Hence the correct answer is Option $(D)$ .

Note:
Here, we need to understand the hidden data given like the initial velocity is zero, the final velocity for first case is uniform velocity for the second case and initial velocity for the third case. Also, by considering the given data and the value we need to find, we need to choose from the four defined equations.