Find the total distance travelled in minimum time taken by car that starts from rest and comes to rest after achieying a peak velocity of 16m/s while moving along a straight line.The magnitude of maximum acceleration and maximum retardation is same at 2m/s^(2).
Asked by anurajdas516
| 27th Dec, 2021,
02:01: PM
Expert Answer:
To get minimum time, acceleration and retardation should be maxiuossible .
Hence we consider acceleration and retrdation as 2 m/s2
Distance S1 travelled while acceleration is determined form the following equation of motion
" v2 = u2 + ( 2 a S1 ) "
where v is velocity attained after travelling a distance S1 with accleration a .
u is initial velocity that is zero because car starts from rest
S1 = v2 / ( 2 a ) = 256 / 4 = 64 m
Distance S2 travelled while acceleration is determined form the following equation of motion
" v2 = u2 - ( 2 a S2 ) "
where v is final velocity after travelling a distance S2 with retardation a .
v is zero because the car comes to rest after travelling distance S2
u = 16 m,/s is initial velocity
S2 = u2 / ( 2 a ) = 256 / 4 = 64 m
Total distance travelled = S1 + S2 = 128 m
Distance S2 travelled while acceleration is determined form the following equation of motion
" v2 = u2 - ( 2 a S2 ) "
where v is final velocity after travelling a distance S2 with retardation a .
v is zero because the car comes to rest after travelling distance S2
u = 16 m,/s is initial velocity
S2 = u2 / ( 2 a ) = 256 / 4 = 64 m
Total distance travelled = S1 + S2 = 128 m
Answered by Thiyagarajan K
| 27th Dec, 2021,
02:49: PM
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