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.
Difficulty: Easy
Solution:
Case – I: (6000 m)
Initial velocity = Vi = $0ms^{-1}$
Time = t = 2 minutes = $2\times 60$ = 120 s
Final velocity = Vf = $48 kmh^{-1}$ = $48 \times \frac{1000}{3600}$ = 13.333 $ms^{-1}$
S1 = $Vav \times t$
S1 =$\frac{(Vf + Vi)}{2}\times$ t
S1 = $13.333 + \frac{0}{2} \times$120
S1 = $6.6665 \times 120$
S1 = 799.99 m = 800 m
Case – II:
Uniform velocity = Vf = $13.333 ms^{-1}$
Time = t = 5 minutes = $5 \times 60$ = 300 s
S2 = $v \times t$
S2 = $13.333 \times$ 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 = $3\times 60$ = 180 s
S3 = $Vav \times$ t
S3 =$\frac{(Vf + Vi)}{2}\times$ 180
S3 = $.333 +\frac{0}{2}\times$ 180
S3 = $6.6665 \times180$
S3 = 1199.97 = 1200 m
Total distance = S = S1 + S2 + S3
S = 800 + 4000 + 1200
S = 6000 m
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