Question

A train is moving at a speed of 90km/hr. Brakes are applied as to produce a uniform acceleration of -0.5m/s². Find how far the train will go before it is brought to rest​

Answer 1

Answer:

Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/s

acceleration , a = -0.5 m/s²

Use formula,

v = u + at

Finally train will be rest so, final velocity, v = 0

0 = 25 - 0.5t

25 = 0.5t ⇒t = 50 sec

Again, use formula,

S = ut + 1/2at²

Where S is distance travelled before stop

S = 25 × 50 - 1/2 × 0.5 × 50²

= 1250 - 1/2 × 0.5 × 2500

= 1250 - 625

= 625 m

Hence, distance travelled = 625m and time taken = 50 sec

Explanation: Hope this helps, please mark me as brainliest

Answer 2

Given:

• Initial speed(u) = 90 km/h
• Acceleration(a) = - 0.5 \: m/s²

To Find:

• Distance travel(s) = ?

Solution:

According to question :

$\sf\mapsto \: Initial \: speed = 90 \times \frac{5}{18} = 25 \: m/s$

$\sf \mapsto \: Final \: speed = 0 \: m/s$

$\sf: \implies {v}^{2} = {u}^{2} + 2as$

$\sf: \implies {0}^{2} = {25}^{2} + 2 \times - 0.5 \times s$

$\sf: \implies 0 = 625 - s$

$\sf: \implies - 625 = - s$

${\bf{ \red{:\implies s = 625 \: m}}}$

Distance travelled by train is 625 m after applying brake.

Additional information:

→ some related formulas

• ${ \bf\: First \: equation \: of \: motion \to \: v = u + at}$
• $\bf \: Second\: equation \: of \: motion \to \: s= u t+ \frac{1}{2} {at}^{2}$
• $\bf \: Time \: of \: flight \to \: t= \frac{ 2u \: {sin} \: \theta }{g}$
• $\bf \: Range \to \: R = \frac{ {u}^{2} sin \: 2 \theta}{g}$
• $\bf \: Maximum \: height \to \: H_{max}= \frac{ {u}^{2} {sin}^{2} \: \theta}{2g}$