.....answer..this.....question..pls...
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sanya55:
pls answer
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#first case#
time = 10 minutes = 1/6 hour
constant velocity = 5 km / hrs
distance (d1) travelled during this time = 5 × 1/6 = 5/6 km
#second case#
deceleration = 1km/hr
I. e, acceleration = -1km/hr
initial velocity, u = 5 km /hr
final velocity , v = 0 km/hr
time , t2 = (v-u) / a
=> t2 = (0-5) / -1 = -5 / -1 = 5
Now v^2 - u^2 = 2as
distance, s =( v^2 - u^2 ) / 2a
=> s =( 0^2 - 5^2 ) /2×-1
=> s = -25 / -2
=> s= 25/2 km
So, total distance covered = d1 + d2 = (5/6 + 25/2) km = ( 5 + 75 ) / 6 km = 80 / 6 km
time = 10 minutes = 1/6 hour
constant velocity = 5 km / hrs
distance (d1) travelled during this time = 5 × 1/6 = 5/6 km
#second case#
deceleration = 1km/hr
I. e, acceleration = -1km/hr
initial velocity, u = 5 km /hr
final velocity , v = 0 km/hr
time , t2 = (v-u) / a
=> t2 = (0-5) / -1 = -5 / -1 = 5
Now v^2 - u^2 = 2as
distance, s =( v^2 - u^2 ) / 2a
=> s =( 0^2 - 5^2 ) /2×-1
=> s = -25 / -2
=> s= 25/2 km
So, total distance covered = d1 + d2 = (5/6 + 25/2) km = ( 5 + 75 ) / 6 km = 80 / 6 km
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