A particle cover 16m during 4th second of its motion and 24m during 6th second of its motion.
Determine the value of initial velocity as well as acceleration.
please give the correct answer otherwise your answer will be reported❤❤❤❤
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A particle cover 16m during 4th second of its motion and 24m during 6th second of its motion.
Determine the value of initial velocity as well as acceleration.
please give the correct answer otherwise your answer will be reported❤❤❤❤
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Verified answer
Answer :
Explanation :
Given :
To find :
Knowledge required :
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Sn = u + ½a(2n - 1)⠀
[Where : Sn = Distance traveled by the body in it's nth second, a = Acceleration of the body, n = nth second, u = Initial velocity of the body]
Solution :
Equation for the motion of the particle in it's 4th s of it's motion :
By using the formula nth second of a particle, we get :
⠀⠀⠀=> Sn = u + ½a(2n - 1)
⠀⠀⠀=> 16 = u + ½a(2(4) - 1)
⠀⠀⠀=> 16 = u + ½a(8 - 1)
⠀⠀⠀=> 16 = u + ½a × 7
⠀⠀⠀=> 32 = 2u + 7a⠀⠀⠀⠀⠀⠀⠀⠀...(i)
Hence the equation for the motion of the particle in it's 4th s of it's motion is (32 = 2u + 7a).
Equation for the motion of the particle in it's 4
6th s of it's motion :
By using the formula nth second of a particle, we get :
⠀⠀⠀=> Sn = u + ½a(2n - 1)
⠀⠀⠀=> 24 = u + ½a(2(6) - 1)
⠀⠀⠀=> 24 = u + ½a(12 - 1)
⠀⠀⠀=> 24 = u + ½a × 11
⠀⠀⠀=> 48 = 2u + 11a⠀⠀⠀⠀⠀⠀⠀⠀...(ii)
Hence the equation for the motion of the particle in it's 6th s of it's motion is (48 = 2u +11a).
Now by subrre5 Eq.(i) and Eq.(ii), we get :
⠀⠀⠀=> 32 - 48 = (2u + 7a) - (2u + 11a)
⠀⠀⠀=> -16 = 2u + 7a - 2u - 11a
⠀⠀⠀=> -16 = -4a
⠀⠀⠀=> 4 = a
⠀⠀⠀⠀∴ a = 4 m/s²
Now by substituting the value of a in the Eq.(i),we get :
⠀⠀⠀=> 32 = 2u + 7a
⠀⠀⠀=> 32 = 2u + 7(4)
⠀⠀⠀=> 32 = 2u + 28
⠀⠀⠀=> 32 - 28 = 2u
⠀⠀⠀=> 2 = u
⠀⠀⠀⠀∴ a = 2 m/s
Therefore,