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Q.

A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates with the  same constant rate and comes to rest. If the total time taken is 4 sec, the distance travelled is

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a

32 m

b

16 m

c

4m

d

None of the above

answer is B.

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Detailed Solution

8=at1 and 0=8−a4−t1 or t1=8a∴8=a4−8a8 =4 a-8 or a=4and t1 = 8/4 = 2secNow, s1=0×2+12×4(2)2 or s1=8ms2=8×2−12×4×(2)2 or s2=8m∴s1+s2=16m
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