Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202415 May 2024Evening ShiftMathematicsDifferentiationActual

After t seconds, the acceleration of a particle, which starts from rest and moves in a straight line is (8- t 5 ) cm / s ^2 , then velocity of the particle at the instant, when the acceleration is zero, is

Options

  1. A160 ~cm / s
  2. B80 ~cm / s
  3. C320 ~cm / s
  4. D480 ~cm / s

Correct answer

A. 160 ~cm / s

Step-by-step solution

Acceleration = (8- t 5 ) cm / s ^2 dv dt =8- t 5 Integrating on both sides, we get v=8 t- t^2 10 +c...(i) At t=0, v=0 array ll & 0=8(0)-0+c c=0 & v=8 t- t^2 10 ...ii[From(i)] array Acceleration =0 aligned & dv dt =0 & 8- t 5 =0 & t =40 aligned Substituting t =40 in (ii), we get Velocity (v)=8(40)- (40)^2 10 =160 ~cm / s

Practice Differentiation on Quantrex Academy →

More from Differentiation

The domain of the derivative of the function f(x)= x 1+|x| is 2025If x= 2^ cosec ⁻¹ t and y= 2^ ⁻¹ t ,|t| 1 then d y d x = 2025If (a+ 2 b x)(a- 2 b y)=a^2-b^2 where a>b>0 , then at ( 4 , 4 ) , dy dx = 2025If x y - y x =0 , then d y d x = 2025If y = ( _ x x )^ x , then dy dx = 2025If f(x)=x^ Sec ⁻¹ x , then f^ (2)= 2025If y= x^4 3 x-5 (x^2-3 )(2 x-3) , then ( d y d x )_ x=2 = 2025If x^2+y^2+ y=4 , then the value of d^2 y d x^2 at x=-2 is 2025 Full Differentiation list All MHT CET PYQs