Acceleration of a particle is given by a=(2+(5 t/2)) m / s 2. At t=0, velocity of the particle is 3 m / s and its position is x=(5/3) m. Find the value of (18/5)((V textavg /V textinst )), where V text avg is average velocity for t=0 to 2 s and V text inst is the instantaneous velocity.at t=2 s.

Question

Acceleration of a particle is given by a=(2+(5 t/2)) m / s 2. At t=0, velocity of the particle is 3 m / s and its position is x=(5/3) m. Find the value of (18/5)((V textavg /V textinst )), where V text avg is average velocity for t=0 to 2 s and V text inst is the instantaneous velocity.at t=2 s. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

V=2 t+(5 t2/4)+3 x=t2+(5 t3/12)+3 t+(5/3) m / s V text avg =(x(2)-x(0)/2)=(20/3) m / s V text inst =V(2)=12 m / s

Q. Acceleration of a particle is given by a=(2+25t​)m/s2. At t=0, velocity of the particle is 3m/s and its position is x=35​m. Find the value of 518​(Vinst​Vavg​​), where V avg ​ is average velocity for t=0 to 2s and Vinst ​ is the instantaneous velocity.at t=2s.

V=2t+45t2​+3 x=t2+125t3​+3t+35​m/s Vavg ​=2x(2)−x(0)​=320​m/s Vinst ​=V(2)=12m/s

$V = 2 t + \frac{5 t ^{2}}{4} + 3$
$x = t^{2} + \frac{5 t ^{3}}{12} + 3 t + \frac{5}{3} m / s$
$V_{avg} = \frac{x ( 2 ) - x ( 0 )}{2} = \frac{20}{3} m / s$
$V_{inst} = V (2) = 12 m / s$