Question

A tram is travelling at a speed
of 90 km hrl. Brakes are applied
so as to produce a uniform
acceleration of -0.5 ms. Find
how far the train will go before it
is brought to rest.
​

Answer 1

Answer:

625 m

Explanation:

apply newton's third eq. of motion

v^2 - u^2 = 2aS

since the train will stop so v=0 put all values and get ans .

whole ans is in pic.

Answer 2

AnswEr:

Given,

• Final Velocity = 0
• Initial velocity = 90 km $\sf{h}^{-1}$

$\tt = \frac{90 \times5 }{18} = 25 \: m \: per \: sec \:$

• Acceleration = 0.5 $\sf{ms}^{-2}$

$\star$ $\sf\underline{Using\:\:third\:\:equation\:\:of\:\:motion:-}$

$\rightarrow \tt {v}^{2} = {u}^{2} + 2as \\ \\ \rightarrow\tt {v}^{2} - {u}^{2} = 2as \\ \\ \rightarrow \tt \frac{ {v}^{2} - {u}^{2} }{2a} = s \\ \\ \rightarrow \tt \: s = \frac{ {v}^{2} - {u}^{2} }{2a} \\ \\ \rightarrow \tt \frac{ {(0)}^{2} - {(25)}^{2} }{2 \times 0.5} = s \\ \\ \rightarrow \tt \: \frac{0 - 625}{ - 1} = s \\ \\ \rightarrow \tt = 625 \: m = s$

Therefore, s = 625 m .