Physics (Kinematics)

Question

A particle moves according to the equation 𝐱 = (𝟓𝒕^𝟑+𝟐𝒕^𝟐 )/ 𝒕 + 𝟖𝒕^𝟐 /𝒕^𝟒 + 𝟒, where x is in meters and t is in seconds. Determine:

(a) The average velocity from 1.5 to 3.0 s.

(b) The instantaneous velocity t = 3.2 s.

(c) The instantaneous acceleration at t = 3.2 s.

(d) The time when the particle is at rest.

Answer 1

A particle moves according to the equation 𝐱 = (𝟓𝒕^𝟑+𝟐𝒕^𝟐 )/ 𝒕 + 𝟖𝒕^𝟐 /𝒕^𝟒 + 𝟒, where x is in meters and t is in seconds. Determine:

(a) The average velocity from 1.5 to 3.0 s.

(b) The instantaneous velocity t = 3.2 s.

(c) The instantaneous acceleration at t = 3.2 s.

(d) The time when the particle is at rest.

As

x = 5t²+2t + 8 t ^-2 +4, X1= 5(1.5)² +2*1.5+8 (1.5) ^-2 +4 and x2 = 5(3*3)² +2*3+8 (3) ^-2 +4

and t2-t1= 3 -1.5= 1.5

(a) For average velocity = total Distance in given time / time taken =( x2 -x1) / 1.5

(b) For velocity dx/dt =d/dt(5t²+2t + 8 t ^-2 +4)= 10t +2 -16 t^-3 +0

Plug in t= 3.2 s to get Velocity at t=3.2

(c) For instant acceleration

a= dv/dt = d/dt (10t + 2 -16t ^-3 )

= 10+ 0 +48 t^-4

Now plug in t= 3.2 s ,get acceleration at t= 3.2 s

( d) for time the particle is at rest v=0

10t +2 -16t ^-3 =0

Solve it for t