Question

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

Answer 1

Answer

• Velocity , v = 67592 m/s
• Acceleration , a = 13506 m/s²

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

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

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