Question

13
A train is travelling at a speed of 90km/hr.Brakes are applied so as to produce a uniform
acceleration of 2m/sec square Find how far the train will go before coming to rest

a 154.25m
b -154.25m
c 156.25m
d -156.25m​

Answer 1

156.25m

Explanation:

train will go before coming to rest is 156.25m

Answer 2

Answer:

Given that,

Acceleration a=2 m/s  

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+2t

25 = 2t

25/2 = t

t = 12.5

Again, using equation of motion,  

$s = ut +\frac{1}{2} at^{2}$

 

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=25×12.5 - 1/2 x 2 x 12.5^2

= 156.25m

this is your required answer

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Explanation: