Question

A train is travelling at a speed of 90 km/h barks are applied so as to produce an unform acceleration of o.5 m/s square. Find how for will the train go before it is brought to rest .

and write and show in note book step by step and you workouted also i will give you a 5 star ​

Answer 1

Answer:

Given that,

Acceleration a=−$0.5m/s^{2}$  

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=  $\frac{25}{0.5}$

t=50 sec

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=25×50−  $\frac{1}{2}$​×0.5×$(50)^{2}$  

s=625 m  

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