Question

if the displacement equation of a moving object is defined as X=3t^4+20t+5 m, where t is in seconds,then find instanteous velocity at 2 second, initial velocity, instanteous acceleration at 7 seconds.​

Answer 1

Given :

➳ Displacement equation of a moving object has been provided.

$\dag\bf\:x=3t^4+20t+5$

To Find :

• Initial velocity
• Instantaneous velocity at 2s
• Instantaneous acc. at 7s

SoluTion :

✴ Instantaneous velocity :

$\leadsto\sf\:v=lim\:(\Delta t\to 0)\:\dfrac{\Delta x}{\Delta t}=\dfrac{dx}{dt}$

$\leadsto\sf\:v=\dfrac{d(3t^4+20t+5)}{dt}$

$\leadsto\bf\:v=12t^3+20$

• Initial velocity (t = 0) :

$\leadsto\sf\:v=12(0)^3+20$

$\leadsto\sf\:v=0+20$

$\leadsto\boxed{\bf{v=20\:ms^{-1}}}$

• Velocity at t = 2s :

$\leadsto\sf\:v=12(2)^3+20$

$\leadsto\sf\:v=96+20$

$\leadsto\boxed{\bf{v=116\:ms^{-1}}}$

✴ Instantaneous acceleration :

$\leadsto\sf\:a=lim\:(\Delta t\to 0)\:\dfrac{\Delta v}{\Delta t}=\dfrac{dv}{dt}$

$\leadsto\sf\:a=\dfrac{d(12t^3+20)}{dt}$

$\leadsto\bf\:a=36t^2$

• Acc. at t = 7s :

$\leadsto\sf\:a=36(7)^2$

$\leadsto\sf\:a=36\times 49$

$\leadsto\boxed{\bf{a=1764\:ms^{-2}}}$

Answer 2

$\blue{\bold{\underline{\underline{Given}}}}$

★ Displacement equation of a moving object is defined as x = 3t⁴ + 20t + 5 m.

$\blue{\bold{\underline{\underline{To \: Find}}}}$

★ Initial velocity

★ Instanteous velocity at 2 second

★ Instanteous acceleration at 7 ssecond

$\blue{\bold{\underline{\underline{Solution}}}}$

↖️Instanteous velocity :

v = lim ( ∆t → 0 ) ∆x/∆t = dx/dt

=> v = d(3t⁴ + 20t + 5)/dt

=> v = 12t³ + 20

✔Initial Velocity (t = 0) :

v = 12(0)³ + 20

=> v = 0 + 20

=> v = 20 ms^-1

✔Velocity at t = 2s :

v = 12(2)³ + 20

=> v = 96 + 20

=> v = 116 ms^-1

↖Instanteous acceleration :

a = lim ( ∆t → 0 ) ∆v/∆t = dv/dt

=> a = d(12t³ + 20)/dt

=> a = 36t²

✔Acceleration at t = 7s :

a = 36(7)²

=> a = 36 × 49

=> a = 1764 ms^-2