1.

PageFind the velocity and acceleration in theequation at t= 15 sec x=5t^4+3t²+2t+1. want answer soon as possible ​

Answer»

Answer

Given

  • Displacement , x = 5t⁴ + 3t² + 2T + 1

To Find

  • Velocity
  • Acceleration

Concept Used

Velocity : It is defined as ratio of displacement and time . ( or )

Velocity over a certain period of time is the instantaneous change in POSITION (dx) over the instantaneous change in time (dt)

\rm \implies v=\dfrac{dx}{dt}

Acceleration : It is defined as ratio of velocity and time ( or )

Acceleration over a certain period of time is the instantaneous change in velocity ( dv ) over the instantaneous change in time ( dt )

\implies \rm a=\dfrac{dv}{dt}

SOLUTION

\rm Velocity,v=\dfrac{dx}{dt}\\\\\implies \rm v=\dfrac{d}{dt}(5t^4+3t^2+2t+1)\\\\\implies \rm v=20t^3+6t+2\\\\\rm Now\ v\ at\ t=15\ s\ is,\\\\\implies \rm v_{15}=20(15)^3+6(15)+2\\\\\implies \rm v_{15}=67500+90+2\\\\\implies \rm v_{15}=67592\ m/s

\rm Acceleration,a=\dfrac{dv}{dt}\\\\\rm \implies a=\dfrac{d}{dt}(20t^3+6t+2)\\\\\implies \rm a=60t^2+6\\\\\implies \rm a_{15}=60(15^2)+6\\\\\implies \rm a_{15}=60(225)+6\\\\\implies \rm a_{15}=13500+6\\\\\implies \rm a_{15}=13506\ m/s^2


Discussion

No Comment Found

Related InterviewSolutions