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CBSE Class 11-science Answered

Car starting from rest accelerates at the rate f through a distance of s and continues at a constant speed for time t and decelerates @ f/2 to come to rest. If the total distance travelled is 5s then prove that s=0.5 ft square.
Asked by tps.mjmdr | 13 Jul, 2018, 07:37: AM
answered-by-expert Expert Answer
speed v attained after traveling distance S with acceleration f is given by,
 
v2 = 2×f×S  ...............(1)
 
distance travelled with uniform speed v = v×t begin mathsize 12px style equals space square root of 2 cross times f cross times S end root space cross times space t end style ..............(2)
final distance Sr travelled with retardation (f/2)  is obtained from  2×f×S  = 2×(f/2)×Sr  
(Aboe eqn. is written using the formula "v2  = u2 - 2×r×S ", with final speed v =0 and initial speed u is to be substituted using eqn.(1). )
Hence Sr = 2×S ............(3)
 
From (1), (2) and (3) we get total distance travelled, S + begin mathsize 12px style space square root of 2 cross times f cross times S end root space cross times space t end style + 2S = 5S (given)
hence begin mathsize 12px style space square root of 2 cross times f cross times S end root space cross times space t end style = 2S  or  S = (1/2) × f × t2 
Answered by Thiyagarajan K | 13 Jul, 2018, 11:01: AM
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