The acceleration of a particle moving only on a horizontal plane is given by a= 3ti +4tj, where a is in meters per second-squared and t is in seconds.

At t = 0s, the position vector r= (20.0 m)i + (40.0 m)j locates the particle, which then has the velocity vector v= (5.000 m/s)i + (2.00 m/s)j

(a) What is the expression of its instantaneous velocity vector, as a function of time?
(b) What is the expression of its instantaneous position vector, as a function of time?

At t = 4s, what are :

(c) its position vector in unit-vector notation?
(d) the angle between its direction of travel and the positive direction of the x axis?

36.5

(a) To find the expression for the instantaneous velocity vector as a function of time, we need to integrate the acceleration vector with respect to time.

Given: a = 3ti + 4tj

Take the integral of each component with respect to time:
∫(3t dt) = (3/2)t^2 + C1,
∫(4t dt) = 2t^2 + C2.

Here, C1 and C2 are integration constants, which we'll determine using the given initial velocity.

Given: v = (5.000 m/s)i + (2.00 m/s)j at t = 0s.

Comparing the velocity components, we get:
(5.000 m/s) = (3/2)(0)^2 + C1,
(2.00 m/s) = 2(0)^2 + C2.

Simplifying, we find C1 = 5.000 m/s and C2 = 2.00 m/s.

Now, substitute the integration constants back into the expressions:
v = (3/2)t^2 + 5.000 m/s i + 2t^2 + 2.00 m/s j.

Therefore, the expression for the instantaneous velocity vector is:
v = (5/2)t^2 + 5.000 m/s i + 2t^2 + 2.00 m/s j.

(b) To find the expression for the instantaneous position vector as a function of time, we need to integrate the velocity vector with respect to time.

Given: v = (5/2)t^2 + 5.000 m/s i + 2t^2 + 2.00 m/s j.

Integrating each component, we get:
∫((5/2)t^2 dt) = (5/6)t^3 + C3,
∫((2t^2 dt) = (2/3)t^3 + C4.

Here, C3 and C4 are integration constants, which we'll determine using the given initial position.

Given: r = (20.0 m)i + (40.0 m)j at t = 0s.

Comparing the position components, we have:
(20.0 m) = (5/6)(0)^3 + C3,
(40.0 m) = (2/3)(0)^3 + C4.

Simplifying, we find C3 = 20.0 m and C4 = 40.0 m.

Now, substitute the integration constants back into the expressions:
r = (5/6)t^3 + 20.0 m i + (2/3)t^3 + 40.0 m j.

Therefore, the expression for the instantaneous position vector is:
r = (11/6)t^3 + 20.0 m i + (2/3)t^3 + 40.0 m j.

(c) To find its position vector at t = 4s, substitute t = 4s into the expression for the position vector:
r = (11/6)(4)^3 + 20.0 m i + (2/3)(4)^3 + 40.0 m j.

Simplifying, we get:
r = 417.333 m i + 170.667 m j.

Therefore, the position vector in unit-vector notation is:
r = 417.333 m i + 170.667 m j.

(d) To find the angle between the direction of travel and the positive x-axis, we need to find the direction of the displacement vector at t = 4s.

Given the position vector r = 417.333 m i + 170.667 m j, we can see that the x-component is much larger than the y-component. This indicates that the particle is moving predominantly in the positive x-direction.

Therefore, the angle between the direction of travel and the positive x-axis is approximately 0 degrees.