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?
(a) To find the expression for the instantaneous velocity vector, we need to integrate the acceleration vector with respect to time:
∫a dt = ∫(3ti + 4tj) dt
Using the power rule of integration, we get:
v = (3/2)t^2 i + (2t^2 + C) j
Since we know the velocity vector v at t = 0s, we can substitute the values and solve for C:
(5.000 m/s)i + (2.00 m/s)j = (3/2)(0)^2 i + (2(0)^2 + C) j
C = 2.00 m/s
Therefore, the expression for the instantaneous velocity vector, as a function of time, is:
v = (3/2)t^2 i + (2t^2 + 2.00 m/s) j
(b) To find the expression for the instantaneous position vector, we need to integrate the velocity vector with respect to time:
∫v dt = ∫((3/2)t^2 i + (2t^2 + 2.00 m/s) j) dt
Using the power rule of integration, we get:
r = (1/2)(3/2)t^3 i + ((2/3)t^3 + 2.00t + C) j
Since we know the position vector r at t = 0s, we can substitute the values and solve for C:
(20.0 m)i + (40.0 m)j = (1/2)(3/2)(0)^3 i + ((2/3)(0)^3 + 2.00(0) + C) j
C = 40.0 m
Therefore, the expression for the instantaneous position vector, as a function of time, is:
r = (1/2)(3/2)t^3 i + ((2/3)t^3 + 2.00t + 40.0 m) j
(c) At t = 4s, to find the position vector in unit-vector notation, we substitute t = 4 into the expression for the position vector:
r = (1/2)(3/2)(4)^3 i + ((2/3)(4)^3 + 2.00(4) + 40.0 m) j
Simplifying and calculating:
r = 27i + 168 j
Hence, the position vector in unit-vector notation at t = 4s is:
r = 27i + 168 j
(d) The angle between the direction of travel and the positive direction of the x-axis can be determined using the formula:
θ = tan^(-1)(vy/vx)
Substituting the given values for the velocity vector v at t = 4s:
θ = tan^(-1)((168)/(27))
Calculating the result:
θ ≈ 80.3°
Therefore, the angle between the direction of travel and the positive direction of the x-axis at t = 4s is approximately 80.3°.