Aparticle's position along the x-axis is described by: x(t)=a t + b t2

where t is in seconds, x is in meters, and the constants a and b are given below.

randomized variables

a = -4.3 m/s

b = 4.9 m/s2

(a) what is the velocity, in meters per second, of the particle at t1 = 3.0 s?
(b) what is the velocity, in meters per second, of the particle when it is at the origin (x=0) at t0 > 0?

The velocity of the particle at t = 3.0s is 25.1 m/s.

The velocity of the particle when it is at the origin is 4.324 m/s.

The given parameters:

The velocity of the particle at t = 3.0s is calculated as follows;

The velocity of the particle when it is at the origin (x = 0)

Learn more about velocity of particle here: link

a)V = 25.1 m/s

b)V = 4.226 m/s

Explanation:

Given that

x(t)=A t + B t²

A = -4.3 m/s

B = 4.9 m/s²

x(t)=  - 4.3 t +4.9 t²

The velocity of the particle is given as

V=-4.3 + 4.9 x 2 t

V= - 4.3 + 9.8  t m/s

Velocity at point t= 3 s

V= - 4. 3 + 9.8 x 3 m/s

V= - 4.3 + 29 .4 m/s

V = 25.1 m/s

At origin :

x= 0 m

0 =  - 4.3 t +4.9 t²

0 = - 4.3 + 4.9 t

t=0.87 s

The velocity at t= 0.87 s

V= - 4.3 + 9.8  t m/s

V= - 4. 3 + 9.8 x 0.87 m/s

V= - 4.3 + 8.526 m/s

V = 4.226 m/s

a)V = 25.1 m/s

b)V = 4.226 m/s

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