Aremote controlled car is moving in a vacant parking lot. the velocity of the car as a function of time is given by v= [5.00 m/s – (0.0180 m/s2)t^2 ]i+[2.00 m/s + (0.550 m/s2)t ]j . what are the magnitude and direction of the velocity of the car at t= 8.00 s?

The magnitude and direction of the velocity of the car at t = 8.00 s is v = 7,468 m / s and the vector direction = θ = 58,984°

In the coordinates of the two-dimensional plane (xy) there is a position vector of a point

You can write P as:

P = xi + yi

where i denotes the unit vector in the x and y directions

While the magnitude of the vector:

and its direction

Likewise with the instantaneous speed which is a derivative of the position function of time

V = vx i + vy j

The velocity of the car as a function of time is given by v = [5.00 m / s - (0.0180 m / s²) t²] i + [2.00 m / s + (0.550 m / s²) t] j so that:

vx = 5 - 0.0180t² i

vy = 2 + 0.55t j

So the instantaneous velocity at t = 8 s becomes:

vx = 5-0.0180.8²

vx = 3,848 m / s

vy = 2+ 0.55.8

vy = 6.4

So the velocity position vector becomes:

v = 3,848 i + 6.4 y

and the magnitude becomes:

v = 7,468 m / s

While the velocity vector direction:

θ = 58,984

Which vector should be negative

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the statements that describe a vector

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the resultant vector

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Keywords: velocity, vector, two-dimensional coordinates,

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