Can you explain the answer of this question below:An athlete uniformly...
v= 80*1000/3600 = 22.22 u=20*1000/3600=5.55 , Now a = v-u/t = 22.22-5.55/2 = 8.335
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Can you explain the answer of this question below:An athlete uniformly...
We use s=ut+1/2at^2 s=200m,. u =0. as it starts from rest. a=1. so,. 200=0×t +1/2×1×t^2 400=t^2 √400=t 20=t
Can you explain the answer of this question below:An athlete uniformly...
Given:
Initial speed, u = 80 km/h = 22.22 m/s
Final speed, v = 20 km/h = 5.56 m/s
Acceleration, a = -2 m/s² (negative sign indicates deceleration)
We need to find the time taken to slow down from 80 km/h to 20 km/h.
Formula:
We can use the formula v = u + at, where
v = final velocity
u = initial velocity
a = acceleration
t = time taken
Rearranging this formula, we get
t = (v-u)/a
Calculation:
Substituting the values in the formula, we get
t = (5.56 - 22.22)/(-2) = 8.33 s (approx)
Therefore, the time taken by the train to slow down from 80 km/h to 20 km/h with a uniform deceleration of 2 m/s² is 8.33 s (approx), which is closest to option C.