Question

A train starts at rest, accelerate with constant acceleration 'a' for 5 minutes, and then decelerates with - a. Suppose it travels a distance of 10km in all. Find a. ​

Answer 1

Answer:

The value of acceleration equals 0.0556m/s^{2}0.0556m/s

2

Explanation:

The distances covered in each of the three phases are calculated as under

1) Distance covered in accelerating phase for a period of 5 minutes or 300 seconds ( 1 minute = 60 seconds)

Using second equation of kinematics

\begin{gathered}s=ut+\frac{1}{2}at^{2}\\\\s_{1}=0\times 300+\frac{1}{2}\times a\times (300)^{2}\\\\s_{1}=0.5\times a\times (300)^{2}\end{gathered}

s=ut+

2

1

at

2

s

1

=0×300+

2

1

×a×(300)

2

s

1

=0.5×a×(300)

2

2) Distance covered in 5 minutes while travelling at a constant speed which is The speed after 5 minutes of travel is obtained by first equation of kinematics as

\begin{gathered}v=u+at\\\\v=0+a\times 300\\\\v=300a\end{gathered}

v=u+at

v=0+a×300

v=300a

Thus distance traveled equals

\begin{gathered}s_{2}=300a\times 300\\\\s_{2}=(300)^{2}a\end{gathered}

s

2

=300a×300

s

2

=(300)

2

a

3)

The phase when the car stops the distance it covers during this phase can be obtained using third equation of kinematics as

\begin{gathered}v^{2}=u^{2}+2as\\\\\therefore s=\frac{v^{2}-u^{2}}{2a}\\\\s_{3}=\frac{0-(300a)^{2}}{2\times -a}\\\\s_{3}=\frac{300^{2}a}{2}\end{gathered}

v

2

=u

2

+2as

∴s=

2a

v

2

−u

2

s

3

=

2×−a

0−(300a)

2

s

3

=

2

300

2

a

Now the sum of s_{1}+s_{2}+s_{3}s

1

+s

2

+s

3

equals 10 kilometers or 10000 meters.

Thus we get

\begin{gathered}0.5\times a\times 300^{2}+300^{2}a+300^{2}\times \frac{a}{2}=10000\\\\a(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})=10000\\\\\therefore a=\frac{10000}{(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})}=0.0556m/s^{2}\end{gathered}

0.5×a×300

2

+300

2

a+300

2

×

2

a

=10000

a(0.5×300

2

+300

2

+0.5×300

2

)=10000

∴a=

(0.5×300

2

+300

2

+0.5×300

2

)

10000

=0.0556m/s

2

Answer 2

The correct answer to the question is 0.11 m/s i.e., a = 0.11 m/s.

Given:

The initial velocity of the train (u) = 0  

(∵ It starts from rest)

The constant acceleration is = a

Time of constant acceleration in the train (t) = 5 minutes

i.e., t = 300 seconds

Total distance covered by the train (s) = 10 Km

i.e., s = 10000 m.

To Find:

The value of a is =?

Solution:

Here, we will use the first and second kinematical equations,

Let the train's final velocity when accelerating be 'v'.

Now, to find the final velocity (v), we will make use of the first kinematical equation which is as follows;

v = u +a(t)

v = 0 + a(300)

∴ v = 300a.

Now, this final velocity will be the initial velocity when the train decelerates and let the time it takes to stop be 'T'.

i.e., u = 300a and v = final velocity = 0

Again, using the first kinematical equation;

v = u + a(t)

0 = 300a + (-a)T

∴ 300a = a(T)

∴ T = 300 seconds

Let, the displacement of the train in acceleration motion and deceleration motion be 's' and 'S' respectively;

Now, by using the second kinematical equations as follows;

s = u(t) + $\frac{1}{2} at^{2}$

∴ s = 0 + $\frac{1}{2}a(300)^{2}$

∴ s =  $\frac{1}{2}a(300)^{2}$                         ----------- (1)

Again,

S = u(T) + $\frac{1}{2}(-a)T^{2}$

∴ S = 300a(300) - $\frac{1}{2}a(300)^{2}$

∴ S =  $\frac{1}{2}a(300)^{2}$                       ------------- (2)

Now, by adding equations (1) and (2), we will get the total displacement of the train,

∴ s + S = 10000

i.e.,   $\frac{1}{2}a(300)^{2}$ +   $\frac{1}{2}a(300)^{2}$  = 10000

∴ $(300)^{2}a$ = 10000

∴ 90000(a) = 10000

∴ a = $\frac{10000}{90000}$

∴ a = 0.11 m/s.

Therefore, the value of 'a' comes out to be 0.11 m/s.

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