Question

A train is moving with a velocity of 54 kmph. It is accelerated at the rate of 5 m/s². Find the distance travelled by the train in 5 seconds and in the 5th second of its journey.​

Answer 1

Explanation:

Given:-

A train is moving with a velocity of 54 kmph. It is accelerated at the rate of 5 m/s².

To Find:-

The distance travelled by the train in 5 seconds and in the 5th second of its journey.

Solution:-

♠ By second equation of motion

$\mapsto \boxed{ \red{ \bf \: s = \: ut + \frac{1}{2}a {t}^{2} }}$

$\rm \: \bigstar \: s = 15 \times 5 + \frac{1}{2} \: \times 5\times {5}^{2} \\ : \longmapsto \: \rm \: s = 72 + 62.5 \\ : \longmapsto \: \rm \: \purple {s = 137.50 \: m}$

→ Now, Distance travelled in 5 sec is 137.50 m.

Now, Distance travelled in th nth sec is given by

$\clubs \mapsto \purple{ \bf \: s_n =u + \frac{1}{2} a(2n - 1)}$

$\rightarrowtail \rm \: s _{5} = \frac{1}{2} \times 5 \times (2 \times 5 - 1) \\ \rightarrowtail \rm \: s _{5} = 15 + 22.5 \\ \rightarrowtail \rm \: \color{maroon} s _{5} = 37.5 \: m$

So, the distance travelled in 5 seconds = 137.50 m and the distance travelled in 5th second is 37.5 m

Answer 2

Explanation:

Given:-

A train is moving with a velocity of 54 kmph. It is accelerated at the rate of 5 m/s².

To Find:-

The distance travelled by the train in 5 seconds and in the 5th second of its journey.

Solution:-

♠ By second equation of motion

\mapsto \boxed{ \red{ \bf \: s = \: ut + \frac{1}{2}a {t}^{2} }}↦

s=ut+

2

1

at

2

\begin{gathered} \rm \: \bigstar \: s = 15 \times 5 + \frac{1}{2} \: \times 5\times {5}^{2} \\ : \longmapsto \: \rm \: s = 72 + 62.5 \\ : \longmapsto \: \rm \: \purple {s = 137.50 \: m}\end{gathered}

★s=15×5+

2

1

×5×5

2

:⟼s=72+62.5

:⟼s=137.50m

→ Now, Distance travelled in 5 sec is 137.50 m.

Now, Distance travelled in th nth sec is given by

\begin{gathered} \clubs \mapsto \purple{ \bf \: s_n =u + \frac{1}{2} a(2n - 1)} \\ \end{gathered}

♣↦s

n

=u+

2

1

a(2n−1)

\begin{gathered} \rightarrowtail \rm \: s _{5} = \frac{1}{2} \times 5 \times (2 \times 5 - 1) \\ \rightarrowtail \rm \: s _{5} = 15 + 22.5 \\ \rightarrowtail \rm \: \color{maroon} s _{5} = 37.5 \: m\end{gathered}

↣s

5

=

2

1

×5×(2×5−1)

↣s

5

=15+22.5

↣s

5

=37.5m

So, the distance travelled in 5 seconds = 137.50 m and the distance travelled in 5th second is 37.5 m