Question

A train accelerates uniformly from 36km/h to 72km/h in 10s. Find the distance travelled

Answer 1

Answer:

Given:

Initial velocity = 36 km/hr

Final Velocity = 72 km/hr

Time taken = 10 seconds

To find:

Distance travelled.

Conversion :

Convert unit of velocity from km/hr to

m/s.

1. 36 km/hr = 36 × 5/18 = 10 m/s

2. 72 km/hr = 72 × 5/18 = 20 m/s

Calculation:

Acceleration be denoted as "a"

$\therefore \: a = \dfrac{v - u}{t}$

$= > a = \dfrac{20 - 10}{10}$

$= > a = 1 \: m {s}^{ - 2}$

Now distance travelled :

$\therefore \: s = ut + \frac{1}{2} a {t}^{2} \\ = > s =( 10 \times 10) + ( \dfrac{1}{2} \times 1 \times {10}^{2} ) \\ = > s = 100 + 50 \\ = > s = 150 \: metres$

So final answer :

$\boxed{ \red{distance = 150 \: metres}}$

Additional information on velocity:

1. It is a vector quantity.

2. It has both directions and magnitude.

3. It is also called the rate of change of Displacement.

Answer 2

$\huge {\red{\boxed{ \overline{ \underline{ \mid\mathfrak{An}{\mathrm{sw}{ \sf{er}} \colon\mid}}}}}}$

Given :

• Initial Velocity (u) = 36 km/h = 10 m/s
• Final Velocity (v) = 72 km/h = 20 m/s
• Time (t) = 10 s

Solution :

Use relation :

$\large \star {\boxed{\sf{v \: = \: u \: + \: at}}} \\ \\ \implies {\sf{20 \: = \: 10 \: + \: a \: \times \: 10}} \\ \\ \implies {\sf{20 \: - \: 10 \: = \: 10a}} \\ \\ \implies {\sf{10a \: = \: 10}} \\ \\ \implies {\sf{a \: = \: \dfrac{10}{10}}} \\ \\ \implies {\sf{a \: = \: 1 \: ms^{-2}}}$

➠ Acceleration is 1 m/s²

______________________________

Use relation :

$\large \star {\boxed{\sf{v^2 \: - \: u^2 \: = \: 2as}}} \\ \\ \implies {\sf{400 \: - \: 100 \: = \: 2s}} \\ \\ \implies {\sf{2s \: = \: 300}} \\ \\ \implies {\sf{s \: = \: \dfrac{300}{2}}} \\ \\ \implies {\sf{s \: = \: 150 \: m}}$

➠ Distance Travelled is 150 m