Question

the displacement of a particle is given by x=18t +5t^2.calculate the instantaneous velocity at t=2s.average velocity between t=2s and t=3s. instantaneous acceleration​

Answer 1

instant . velocity =dx/dt

x=18t+5t^2

d (18t+5t^2)/dt

t=1

t^2=2t

18×1+5×2t

18+10t - eq1

at t=2s

v= 18+10t

18+10×2

18+20

38m/s

Ave velocity =x1-x2/t1-t2

t2=3s t1=2s

x2 =18t+5t^2s

18×3+5×9=99

x1=18t+5t^2

18×2+5×4=56

Ave velocity =99-56/3-2=53m/s

instantaneous acceleration =dv/dt

v=18 +10t from eq.1

d (18+10t )/dt

18=0 (constant )

t=1

= 0+10×1

10 m/s^2

Answer 2

Answer :

Displacement = x = 18t + 5t².

Differentiating x :

⇒ v = dx/dt

⇒ 18 + 2*5t

⇒ 18 + 10t

Now, Finding the Instantaneous velocity at 2 seconds :

⇒ 18 + 10*2

⇒ 18 + 20

⇒ 38 m/s

Hence, Instantaneous velocity at time of 2 seconds = 38 m/s.

Now, Finding the Instantaneous velocity at 2 seconds :

⇒ 18 + 10*3

⇒ 18 + 30

⇒ 48 m/s

Hence, Instantaneous velocity at time of 3 seconds = 48 m/s.

Now, Finding Average Velocity :

⇒ (v2 - V1)/(t2 - t1)

⇒ (48 - 38)/(3 - 2)

⇒ 10/1

⇒ 10 m/s

Hence, Average Velocity between 3 s and 2 s is 10 m/s.

Now, Differentiating v :

⇒ a = dv/dt

⇒ 0 + 10

⇒ 10 m/s²

Hence, The Instantaneous acceleration is 10 m/s².