Question

a train is travelling at a speed of 90 km/h brakes are applied so as to produce an uniform acceleration of - 10m/s^2. Find how far the train will go before its brought to rest
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Answer 1

Answer:

625 m

Step-by-step explanation:

Given that,

Acceleration a=-10

2

Speed v=90km/h=25m/s

Using equation of motion,

v=u+at

Where,

v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the equation

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

0=25−0.5t

25=0.5t

t=

0.5

25

t=50 sec

Again, using equation of motion,

S=ut+

2

1

at

2

Where, s = distance

v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the equation

Where S is distance travelled before stop

s=25×50−

2

1

×0.5×(50)

2

s=625 m

So, the train will go before it is brought to rest is 625 m.

Answer 2

Answer:

31.25 m

Step-by-step explanation:

Initial velocity of the train( u):

90 km/h

ON CONVERTING IT TO m/s:

$90 \times \frac{5}{18} = 25 \: m {s}^{ - 1}$

ACCELERATION OF TRAIN( a):

- 10 m/s²

FINAL VELOCITY(v) ⇒ 0 m/s²

WE KNOW THAT ,

⇒ v²= u² + 2 as

$s \: = \frac{ {v}^{2} - {u}^{2} }{2a}$

TOTAL DISTANCE COVERED BEFORE COMING TO REST (s):

$\frac{ {0}^{2} - {25}^{2} }{2 \times( - 10)} = \frac{ - 625}{ - 20} = 31.25 \: m$

DISTANCE COVERED BEFORE COMING TO REST= 31.25 m