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an automobile can accelerate or decelarate at a maximum value of 5 / 3 m/s^2 and can attain a maximum speed of 90 km / hr . if it starts from rest , what is the shortest time in which can travel one kilometre . if it is to come to rest at the end of the kilometer run ?
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Answered by
200
answer :-
in order that the time of motion be shortest , the car should attain the maximum acceleration after the start , maintain the maximum velocity for as along as possible and then decelerate with the maximum retardation possible , consistent with the condition that , the automobile should come to rest immediately after covering a distance of 1 km.
Let T1 be the time of acceleration , T2 be the time of retardation .
now , maximum velocity possible = 90 km /hr =>
90 × 5 /18 m/s = 25m/s
let T1 = v - u / a = 25 - 0 / 5 /3 = 15s
similarly , the me of retardation is also given by ,
T3 = 0 - 25 / 5 / -3 = 15 s
= 15s
during the period of acceleration , the distance covered = average velocity × time = 25 + 0 / 2 = 187 . 5 m
during period of deaccelarattion , the distance covered is the same , hence,= 187.5 m
the total distance covered under constant velocity = 1000 - 375 = 625 m
time of motion under constant velocity T2 = 625 / 25 = 25s
the shortest time of motion = T1 + T2 + T3
= 15 + 25 + 15
= 55second.
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in order that the time of motion be shortest , the car should attain the maximum acceleration after the start , maintain the maximum velocity for as along as possible and then decelerate with the maximum retardation possible , consistent with the condition that , the automobile should come to rest immediately after covering a distance of 1 km.
Let T1 be the time of acceleration , T2 be the time of retardation .
now , maximum velocity possible = 90 km /hr =>
90 × 5 /18 m/s = 25m/s
let T1 = v - u / a = 25 - 0 / 5 /3 = 15s
similarly , the me of retardation is also given by ,
T3 = 0 - 25 / 5 / -3 = 15 s
= 15s
during the period of acceleration , the distance covered = average velocity × time = 25 + 0 / 2 = 187 . 5 m
during period of deaccelarattion , the distance covered is the same , hence,= 187.5 m
the total distance covered under constant velocity = 1000 - 375 = 625 m
time of motion under constant velocity T2 = 625 / 25 = 25s
the shortest time of motion = T1 + T2 + T3
= 15 + 25 + 15
= 55second.
mark me as brain list.
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nice explanation.. well written
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Answered by
49
hello friend nice question and little bit interesting
let T1 be the time of acceleration
late T2 to be the time of retardation
so,
velocity possible is equal to 90 km per hour => 90*5/18m/s=25m/s
let T1 equal to v-u/a=25-0/5/3=15s
then, time of retardation is given as
T3=0-25/5/(-3)=15s
during acceleration the distance
= average velocity ×time
=25+0/2=187.5m
during acceleration the distance covered is same.
Total distance equal 1000 - 375 =625
T2 is equal to 625 by 25 that equal to 25 s
shortest time=
T1+T2+T3= 15+25+15=55s
is ur ans.
let T1 be the time of acceleration
late T2 to be the time of retardation
so,
velocity possible is equal to 90 km per hour => 90*5/18m/s=25m/s
let T1 equal to v-u/a=25-0/5/3=15s
then, time of retardation is given as
T3=0-25/5/(-3)=15s
during acceleration the distance
= average velocity ×time
=25+0/2=187.5m
during acceleration the distance covered is same.
Total distance equal 1000 - 375 =625
T2 is equal to 625 by 25 that equal to 25 s
shortest time=
T1+T2+T3= 15+25+15=55s
is ur ans.
nice ans. !! nice try...!!
thanks a lot sister
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