A train starting from rest accelerates uniformly and attains a velocity

Question

A train starting from rest accelerates uniformly and attains a velocity of 48 $kmh^{-1}$ in 2 minutes. It travels at this speed for 5 minutes. Finally, it moves with uniform retardation and is stopped after 3 minutes. Find the total distance traveled by train.

Accepted Answer

Solution:

Case – I: (6000 m)

Initial velocity = Vi = $0ms^{-1}$

Time = t = 2 minutes = $2 imes 60$ = 120 s

Final velocity = Vf = $48 kmh^{-1}$ = $48 imes \frac{1000}{3600}$ = 13.333 $ms^{-1}$

S1 = $Vav imes t$

S1 =$\frac{(Vf + Vi)}{2} imes$ t

S1 = $13.333 + \frac{0}{2} imes$120

S1 = $6.6665 imes 120$

S1 = 799.99 m = 800 m

Case – II:

Uniform velocity = Vf = $13.333 ms^{-1}$

Time = t = 5 minutes = $5 imes 60$ = 300 s

S2 = $v imes t$

S2 = $13.333 imes$ 300

S2 = 3999.9 = 4000 m

Case – III:

Initial velocity = Vf = $13.333 ms^{-1}$

Final velocity = Vi = 0 $ms^{-1}$

Time = t = 3 minutes = $3 imes 60$ = 180 s

S3 = $Vav imes$ t

S3 =$\frac{(Vf + Vi)}{2} imes$ 180

S3 = $.333 +\frac{0}{2} imes$ 180

S3 = $6.6665 imes180$

S3 = 1199.97 = 1200 m

Total distance = S = S1 + S2 + S3

S = 800 + 4000 + 1200

S = 6000 m