A remote-controlled car is moving in a vacant parking lot. The velocity of the c

Question

A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by [5.00 m/s(0.0180 m/s^3) t^2 ] i^ + [2.00m/s+(0.550m/s^2)t] j^.

(a) What is the magnitude of the velocity of the car at t = 6.07 s ?

(b) What is the direction (in degrees counterclockwise from +x-axis) of the velocity of the car at t = 6.07 s ?

(c) What is the magnitude of the acceleration of the car at t = 6.07 s ?

(d) What is the direction (in degrees counterclockwise from +x-axis) of the acceleration of the car at t = 6.07 s ?

Explanation / Answer

The equation in your question has been 'chopped off'; it says:
[5.00m/s-(.0180m/s^3)t^2]i(hat)+[2.00m/s+(0.550m/s^2)t] j^

For illustration, suppose (missing out units, for clarity) the equation is:
v = (5 -.0180t²)î + (2 + 5t)

Also, I'm guessing '6.07' should be '6.07s'.

Key facts you need to know:

The magnitude of the vector aî + b is (a² + b²).

The direction of the vector aî + b is tan¹(b/a) when measured anticlockwise from the +x axis.
You have to consider the signs of a and b to make sure you are in the correct quadrant.

So for illustration, here's the method in summary.
______________________________________

A) When t=6.07s
v = (5 - 0.018x6.07²)î + (2 + 5x6.07)

So the magnitude of v is
||v|| = [(5 - 0.018x6.07²)² + (2 + 5x6.07)²] (units are m/s)

=32.63 m/s
______________________________________

B) The direction is tan¹ [(2 + 5x6.07) / (5 - 0.018x6.07²) ]

=82.36
___________________________

C) a = dv/dt

v = (5 - 0.0180t²)î + (2 + 5t)
a = dv/dt = (2 x 0.018t)î + 5

When t=6.07s
a = (2 x 0.018x6.07)î + 5

||a|| = [(2 x 0.018x6.07)² + 5²]

=5.004
_____________________

D) The direction is tan¹ [5/(2 x 0.018x6.07)]

=0.422

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