Question

A train is travelling at a speed of 90Km/hr. Brakes are applied so as to produce an
uniform acceleration of 10m/s2. Find how far the train will go before it is brought to rest.


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Answer 1

Answer:

$\huge\underline\mathrm\purple {Answer⬇}$

Acceleration a=−0.5m/s 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=

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.

Hence, A is correct.

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

$\huge\mathtt \red{A} \blue{n} \pink{s} \purple{w} \blue{e} \orange{r}$

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= 0.525

t=50 sec

Again, using equation of motion,

S=ut+ 21 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− 21 ×0.5×(50) 2

s=625 m

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

Hence, A is correct.

I hope it is helpful for you

@ Aman jha