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 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 ?

Answer 1

Equation of velocity is given as

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

at t = 7.93 s

$v = 3.87 \hat i + 6.36 \hat j$

so the magnitude of the velocity is given as

$v = \sqrt{3.87^2 + 6.36^2}$

$v = 7.45 m/s$

Part b)

the direction of the velocity is given as

$\theta = tan^{-1}\frac{6.36}{3.87}$

$\theta = 58.7 degree$

part c)

for acceleration we know that

$a = \frac{dv}{dt}$

$a = -0.036 t\hat i + 0.550\hat j$

at t = 7.93 s

$a = -0.285\hat i + 0.550\hat j$

magnitude is given as

$a = \sqrt{0.285^2 + 0.550^2}$

$a = 0.62 m/s^2$

Part d)

for the direction of the motion

$\theta = tan^{-1}\frac{0.550}{-0.285}$

$\theta = 117.4 degree$