A train accelerating uniformly from rest attains a maximum speed of 40ms−1 in 20s. It travels at this speed for 20s and is brought to rest by uniform retardation in further 40s. What is the magnitude average velocity during this period?
A
80/3ms−1
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B
40ms−1
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C
25ms−1
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D
30ms−1
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Solution
The correct option is C25ms−1 Using first equation of motion v=u+at 40=(20)a1(∵u=0) ∴a1=2m/s2 ⇒S1=12a1t2 ∴S1=400m
Thereafter it travels with constant speed of 40m/s, hence distance travelled will be, ⇒S2=40×20=800m
Further it travels with uniform retardation until final velocity becomes 0 Putting t=40s,u=40m/s and v=0 in first equation of motion, we get 0=40+(a2)t −40=(40)a2 ∴a2=−1m/s2 ∴ retardation = 1m/s2
Putting a=−1m/s2 in third equation of motion, we get 02−402=2(−1)S3 ⇒S3=800m
Now average velocity =Total displacementtime =400+800+80020+20+40=25m/s