Question

(b) A train starting from rest attains a velocity of 90 km/h in 8 minutes. Assuming that
the acceleration is uniform, find out the
(i) acceleration
(ii) the distance travelled by the train for attaining this velocity.
Pls answer as fast as possible​

Answer 1

Given:-

☞Initial velocity=0

☞Final Velocity  = 90 km/h

☞time taken = 8min

 

 

Find:-

☞(i) acceleration

☞(ii) the distance travelled by the train for attaining this velocity.

 

 

Formula to be used:-

☞ $\boxed{\bf{v = u + at}}$

☞ $\boxed{\bf{s = ut + \dfrac{1}{2 } at^2 }}$

Where,

v=final velocity,

u= initial velocity,

a= acceleration,

t=time,

s=distance.

Solution:-

☞First change time min. to sec.

we know,

1min= 60 sec

8 min = 8×60 sec

Time taken = 480 sec

 

☞Now change Velocity km/h to m/s

we know,

1km= 1000m

1h= 3600 sec

$\sf{1km/h = \dfrac{1000}{3600} m/s= \dfrac{5}{18}m/s }$

so,

$\sf{ 90km/h = \dfrac{5}{18} \times 90 m/s = 25m/s }$

Final velocity = 25m/s

 

☞i)Acceleration:-

By using 1st equation of motion,

v=u+at

put values,

$\sf{\implies 25 = 0+ a \times 480}$

$\sf{\implies a= \dfrac{25}{480} }$

So, acceleration = 0.0520m/s²

 

☞ii)the distance travelled by the train for attaining this velocity.

By using 2nd equation of motion,

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

put values,

$\sf{s= 0\times 480 + \dfrac{1}{2} \times 0.0520 \times 480^2}$

$\sf{ s= 0.026 \times 230400}$

So, the distance covered is 5990.4m.

Answer 2

Answer:

GIVEN:

A TRAIN STARTING FROM REST ATTAINS A VELOCITY OH 90 KM/H IN 8 MINS AND ACCELERATION IS UNIFORM.

TO FIND:

1.ACCELERATION

2. THE DISTANCE TRAVELLED BY THE TRAIN FOR ATTAINING THIS VELOCITY.

FORMULA TO BE USED:

$a = \frac{v - u}{t} \\ \\ {v}^{2} = {u}^{2} + 2as$

WHERE,

a=ACCELERATION

v=FINAL VELOCITY

u=INITIAL VELOCITY

t=TIME TAKEN

s=DISTANCE TRAVELLED

SOLUTION:

V= 90 KM/H =90×5/18=25 M/S

U= 0 M/S

TIME TAKEN = 8 MIN=8×60=480 SECONDS

A=V-U/T=90-0/480=90/480=1/4=0.25 M/S^2

NOW DISTANCE TRAVELLED:

90^2=0^2+2×1/4×S

=>8100=0+1/2×S

=>8100=1/2×S

=>8100×2=S

=>S=16200 M.

HENCE ,

ACCELERATION =0.25 M/S^2

DISTANCE TRAVELLED =16200 M.