A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by V =[5.00m/s−(0.0180m/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 = 7.93 s ?
B) What is the direction (in degrees counterclockwise from +x-axis) of the velocity of the car at t = 7.93 s ?
C) What is the magnitude of the acceleration of the car at t = 7.93 s ?
D) What is the direction (in degrees counterclockwise from +x-axis) of the acceleration of the car at t = 7.93 s ?

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aristocles aristocles · Answer 1 · 2018-09-20T03:45:28+02:00

Equation of velocity is given as

[tex]V =[5.00m/s−(0.0180m/s^3)t^2]i^ + [2.00m/s+(0.550m/s^2)t]j^[/tex]

at t = 7.93 s

[tex]v = 3.87 \hat i + 6.36 \hat j[/tex]

so the magnitude of the velocity is given as

[tex]v = \sqrt{3.87^2 + 6.36^2}[/tex]

[tex]v = 7.45 m/s[/tex]

Part b)

the direction of the velocity is given as

[tex]\theta = tan^{-1}\frac{6.36}{3.87}[/tex]

[tex]\theta = 58.7 degree[/tex]

part c)

for acceleration we know that

[tex]a = \frac{dv}{dt}[/tex]

[tex]a = -0.036 t\hat i + 0.550\hat j[/tex]

at t = 7.93 s

[tex]a = -0.285\hat i + 0.550\hat j[/tex]

magnitude is given as

[tex]a = \sqrt{0.285^2 + 0.550^2}[/tex]

[tex]a = 0.62 m/s^2[/tex]

Part d)

for the direction of the motion

[tex]\theta = tan^{-1}\frac{0.550}{-0.285}[/tex]

[tex]\theta = 117.4 degree[/tex]