Question

make thus qyestion using 2nd equation of motion.​

Answer 1

u = 90*5/18 = 25 m/s

v = 0

a = -0.5

t= ?

v = u + at

0= 25 -0.5t

-25 = -0.5t

t= 25/0.5

t = 50 seconds

Second equation of motion

s = ut + 1/2 at²

s = 25*50 + 1/2 * 0.5 *50 *50

s = 1250 + 1/2 * 2,500

s = 1250 + 1250

s = 2500

distance travelled = 2500 metre or 2.5 km

Answer 2

Answer :

Given Initial speed of train (u) =90 km h-1

= (90 1000) / (60×60) m s-1

= 25 m s-1

Final speed of train (v) = 0 ms1

Braking acceleration (a) = -0.5 m s-2

We know 2as = v2- u2

Or distance (s) =(v2-u2)/2a

∴ Distance covered by the train before it came to rest =(02-252)/(2 ×-0.5 )m

= - (25 × 25)×10/-1 m= 625 ans