A lift ascends from rest with uniform acceleration of 4m/s^2, then it moves with uniform velocity and finally comes to rest with uniform retardation of 4m/s^2. If the total distance covered during ascending is 28m and total time taken for ascending is 8s respectively. Find the time for which the lift moves with uniform velocity .Also find its uniform Velocity.
Answers
Case 1 : ascending with uniform accrltn
initial velocity u = 0m/s , a= 4m/s² , time = t₁ sec , final velocity =v m/s
distance covered in t₁ = s₁ m
v₁ = 0 + 4t₁ , 2xs₁x4 + 0² = v² ⇒ s₁=v²/8
case 2 : moving with uniform velocity v
since uniform velocity therefore no acceleration
initial velocity = v m/s , a = 0 m/s² , time = t sec , distance in t = s₂ m final velocity = v m/s
s₂ = vt
case3 : ascending with retardation
initial velocity = v m/s , time taken = t₂ sec , distance travelled = s₃ m
accleration = -4 m/s² final velocity = 0m/s
2x -4 x s₃ + v² = 0² ⇒ s₃= v²/8 , 0 = v -4t₂ ⇒ t₂ = v/4
now A/Q
total time = t₁ + t + t₂ = 8 ⇒ v/4 + t + v/4 ⇒ t = 8 - v/2..........(1)
total distance = s₁+ s₂ + s₃ = 28 ⇒ v²/8 + v²/8 + vt = 28 ⇒ v²/4 + 4vt = 112
⇒substituting value of t from (1) gives quadratic equation in terms of v
v² - 32v + 112 = 0
on solving we get v= 4m/s and v= 28m/s
t = 8- 4/2 = 6 sec and t = 8-28/2 = -6 sec as time can't be negative
true ans is v= 4m/s and t = 6 secs
hope you'll get it, if any queries please ask ........
Answer:v=4s
Explanation:Brainly.in
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Secondary School Physics 5 points
A lift ascends from rest with uniform acceleration of 4m/s^2, then it moves with uniform velocity and finally comes to rest with uniform retardation of 4m/s^2. If the total distance covered during ascending is 28m and total time taken for ascending is 8s respectively. Find the time for which the lift moves with uniform velocity .Also find its uniform Velocity.
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parisakura98pari Ace
There r 3 cases
Case 1 : ascending with uniform accrltn
initial velocity u = 0m/s , a= 4m/s² , time = t₁ sec , final velocity =v m/s
distance covered in t₁ = s₁ m
v₁ = 0 + 4t₁ , 2xs₁x4 + 0² = v² ⇒ s₁=v²/8
case 2 : moving with uniform velocity v
since uniform velocity therefore no acceleration
initial velocity = v m/s , a = 0 m/s² , time = t sec , distance in t = s₂ m final velocity = v m/s
s₂ = vt
case3 : ascending with retardation
initial velocity = v m/s , time taken = t₂ sec , distance travelled = s₃ m
accleration = -4 m/s² final velocity = 0m/s
2x -4 x s₃ + v² = 0² ⇒ s₃= v²/8 , 0 = v -4t₂ ⇒ t₂ = v/4
now A/Q
total time = t₁ + t + t₂ = 8 ⇒ v/4 + t + v/4 ⇒ t = 8 - v/2..........(1)
total distance = s₁+ s₂ + s₃ = 28 ⇒ v²/8 + v²/8 + vt = 28 ⇒ v²/4 + 4vt = 112
⇒substituting value of t from (1) gives quadratic equation in terms of v
v² - 32v + 112 = 0
on solving we get v= 4m/s and v= 28m/s
t = 8- 4/2 = 6 sec and t = 8-28/2 = -6 sec as time can't be negative
true ans is v= 4m/s and t = 6 secs