Question

A subway train starts from rest at a station and accelerates at a rate of 1.95 m/s² for 15.0 s. It runs at constant speed for 74.0 s and slows down at a rate of 3.20 m/s² until it stops at the next station. Find the total distance covered.

		2.51 km
		2.52 km
		2.53 km
		2.56 km

Answer 1

disatnce travelled during first 15 secs is S1 = 0.5 at^2

S1 = 0.5 * 1.95 * 15*15

S1 = 210.375 m ------------------------------------------1

spped acquired during this time is V = u +a t

V = 0 + 15* 1.95

V = 29.25 m/s

distance travelle at this spped for 74 secs

S2 = ut = 29.25* 74   = 2164. 5m ----------------------2

distannce travelled during with acccelration 3.2 m/s^2 is V^2 = 2aS

S3 = (29.25*29.25/(2*3.2)

S3 = 133.68 m

so total distance S = s1+S2+s3

S = 210.375 + 2164.5 + 133.68

S = 2.51 Km ------<<<<<<<<<<<<,answer

opyion A is Correct

Answer 2

acceleration phase

s=1/2*at^2=0.5*1.95*15^2=219.375 m

constant velcoity phase

v=at=15*1.95=29.25 m/s

s=vt=29.25*74=2164.5 m

deceleration phase

v=u-at=29.25-3.2*t=0

t=9.14 s

s=ut-1/2*at^2=29.25(9.14)-0.5*3.2*(9.14)^2=133.68 m

Total=133.68+219.375+2164.5=2517.55 m=2.517 km

Answer 3

distance travelled during acceleartion=at²/2=1.95*15²/2=219.375 m

velocity after acceleration=v=u+at=0+at=1.95*15=29.25 m/s

distance travelled during constant velocity=vt=29.25*74=2164.5 m

distance travelled durind decceleration=(v²-u²/2a)=29.25²/2*3.20=133.68

total distance=219.375+2164.5+133.68=2517.555=2.52 km