Question

A train accelerates uniformly from 18 km h-' to
108 km h'in 25 s. Calculate the
(a) acceleration.
[Ans. I ms)
(b) distance covered.
(Ans. 437.5 m]​

Answer 1

Answer:

Initial speed u = 18 km/hr = 18 × (5/18) = 5 m/s ;Final speed v = 108 km/hr = 108 × (5/18) = 30 m/s

acceleration a is calculated using the formula " v = u+at " , where t is time in seconds.

acceleration a = (v-u)/t = (30-5)/25 = 1 m/s2

Distance S is calculated using the equation S = u×t + (1/2)×a×t2 = 5×25 + (1/2)×1×25×25 = 437.5 m

Answer 2

Answer:

Given :-

• A train accelerates uniformly from 18 km/h to 108 km/h in 25 seconds.

To Find :-

• Acceleration
• Distance Covered

Formula Used :-

$\clubsuit$ First Equation Of Motion :

$\longmapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

$\clubsuit$ Second Equation Of Motion :

$\longmapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}at^2}}}$

where,

• s = Distance Covered
• u = Initial Velocity
• v = Final Velocity
• a = Acceleration
• t = Time

Solution :-

First, we have to convert km/h to m/s :

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: case\: of\: initial\: velocity\: :-}}}}}$

$\implies \sf Initial\: Velocity =\: 18\: km/h$

$\implies \sf Initial\: Velocity =\: {\cancel{18}} \times \dfrac{5}{\cancel{18}}\: m/s$

$\implies \sf\bold{\green{Initial\: Velocity =\: 5\: m/s}}$

${\normalsize{\bold{\purple{\underline{\bigstar\: In\: case\: of\: final\: velocity\: :-}}}}}$

$\implies \sf Final\: Velocity =\: 108\: km/h$

$\implies \sf Final\: Velocity =\: 108 \times \dfrac{5}{18}\: m/s$

$\implies \sf Final\: Velocity =\: \dfrac{540}{18}\: m/s$

$\implies \sf\bold{\green{Final\: Velocity =\: 30\: m/s}}$

Now, we have to find the acceleration :

Given :

• Initial Velocity (u) = 5 m/s
• Final Velocity (v) = 30 m/s
• Time (t) = 25 seconds

According to the question by using the formula we get,

$\longrightarrow \sf v =\: u + at$

$\longrightarrow \sf 30 =\: 5 + a(25)$

$\longrightarrow \sf 30 - 5 =\: 25a$

$\longrightarrow \sf 25 =\: 25a$

$\longrightarrow \sf \dfrac{\cancel{25}}{\cancel{25}} =\: a$

$\longrightarrow \sf\bold{\red{a =\: 1\: m/s^2}}$

$\therefore$ The acceleration of a train is 1 m/s².

Now, we have to find the distance covered :

Given :

• Initial Velocity (u) = 5 m/s
• Time (t) = 25 seconds
• Acceleration (a) = 1 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf s =\: 5 \times 25 + \dfrac{1}{2} \times 1 \times (25)^2$

$\longrightarrow \sf s =\: 125 + \dfrac{1}{2} \times 1 \times 625$

$\longrightarrow \sf s =\: 125 + \dfrac{1}{\cancel{2}} \times {\cancel{625}}$

$\longrightarrow \sf s =\: 125 + 312.5$

$\longrightarrow \sf\bold{\red{ s =\: 437.5\: m}}$

$\therefore$ The distance covered by the train is 437.5 m .