A train, moving at 25 m/s, is brought to rest in 50 sec. Find the acceleration and the time taken.?

Question

u=initial speed(speed at the start)

v=final speed

a=acceleration

t=time

s=displacement(distance travelled)

formulas:

v= u + at

s= ut + 1/2atsquared

vsquared=usquared + 2as

you can use them if u want but u can do ur own way...thankyou in advance..i nid it asap..pls2..help!..

Answer 1

Every 2 seconds, the train slows down by, 1m.s

In 50 seconds, the train would be at a compleate halt...

Very Easy...

You could slow it down even faster...

Every 1 second, the train slows down by, 1m.s

It would come to a compleate stop in half the time... 25s

Answer 2

u = 25

v = 0

t = 50

v = u+at

therefore a = (v - u)/50 = -25/50 = -1/2m/s^2

therefore deceleration = 1/2 m/s^2

s = ut + 1/2 at^2

= 0 + 1/2 * 1/2 *2500 = 625m

Answer 3

decelleration is 1/2 meters per sec and distance travelled is 625m

applied maths...oh the joys:P

Answer 4

u=25m/s

v=0m/s

t=50sec

v=u+.at

putting the values,

0=25.+a*50

=>-25/50=a;

=>a=-1/2;

as a -ve therefore it is retardation.

to find distance

s=ut+.1/2at^2;

by putting the values

s=25*50+.1/2*(-1/2)*50*50;

=>s=1250-625;

=>s=625;

therefore distance covered is 625m and retardation is 0.5m/s^2

Answer 5

First formula to use is

Vf - Vo = aT

where

Vf = final velocity = 0 (when the train stops)

Vo = initial velocity = 25 m/sec (given)

a = acceleration

T = time = 50 sec.

Substituting appropriate values,

0 - (25) = (a)(50)

a = -25/ 50

a = -0.50 m/sec^2

NOTE -- the negative sign indicates that the train was decelerating during this interval.

The next formula to use is

Vf^2 - Vo^2 = 2aS

where

S = distance travelled by train before coming to a stop

and all the other terms have been previously defined.

Substituting values,

0 - (25)^2 = (2)(-0.5)(S)

Solving for "S",

S = -625/(-1)

S = 625 meters  

Answer 6

u= 25m/s

v=0m/s(brought to rest)

t=50 s

so using

v=u+at

0=25+50a

-25=50a

-0.5=a

(negative sign indicates acceleration is opposite in direction to velocity or Deceleration)

therefore, deceleration, a= 0.5m/s^2 or acceleration ,a =-0.5 m/s^2

now, use either of the other 2 formulas to find distance covered..

s=ut+1/2(at^2)

s=25*50+1/2(-0.5)*(50*50)

s=1250-625

s=625 m

or,

v^2 =u^2 +2as

0*0=25*25 +2*(-0.5)*s

625=s

therefore,

distance covered , s= 625 m

hope dis helps.. :)