A particle starts with initial speed u and retardation a to come to rest in time T. The time taken to cover first half of the total path travelled is ????? give answer in terms of T.
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Draw the velocity-time graph for the motion. It will be a straight line corresponding to
v = u - at
at t = T, v = 0
Thus
u = aT
a = u/T (slope of the line)
The line will intercept y axis at u and x-axis at T
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Total distance travelled S = Area under the v-t graph
S = 1/2 u x T = uT/2
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Then half distance S/2 = uT/4
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Let the time taken to complete S/2 be t
S/2 = ut - 1/2 at2
uT/4 = ut - 1/2 (u/T) t2
T/4 = t - 1/2 t2/T
It will become a quadratic
t2 - 2Tt + T2/2 = 0
t2 - 2Tt + T2 = T2 /2
(t - T)2 = T2/2
t - T = + - T/√2
t = T +- T/√2
We take the negative sign because with positive sign the value will exceed T which is not possible.
Hence t = T - T/√2 = (2 - √2) T/2
t = 0.293 T
t = 29.3 % of total time
v = u - at
at t = T, v = 0
Thus
u = aT
a = u/T (slope of the line)
The line will intercept y axis at u and x-axis at T
-
Total distance travelled S = Area under the v-t graph
S = 1/2 u x T = uT/2
-
Then half distance S/2 = uT/4
-
Let the time taken to complete S/2 be t
S/2 = ut - 1/2 at2
uT/4 = ut - 1/2 (u/T) t2
T/4 = t - 1/2 t2/T
It will become a quadratic
t2 - 2Tt + T2/2 = 0
t2 - 2Tt + T2 = T2 /2
(t - T)2 = T2/2
t - T = + - T/√2
t = T +- T/√2
We take the negative sign because with positive sign the value will exceed T which is not possible.
Hence t = T - T/√2 = (2 - √2) T/2
t = 0.293 T
t = 29.3 % of total time
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