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

A remote-controlled car is moving in a vacant parking lot with a given velocity function. The magnitude of the velocity at t = 7.93 s is 8.05 m/s, and the direction is -60.6 degrees counterclockwise from the +x-axis. The magnitude of the acceleration at t = 7.93 s is 0.616 m/s^2, and the direction is -63.7 degrees counterclockwise from the +x-axis.

Step-by-step explanation:

A) What is the magnitude of the velocity of the car at t = 7.93 s?

To find the magnitude of the velocity, we need to calculate the magnitude of V at t = 7.93 s. Using the given equation V =[5.00m/s−(0.0180m/s^3)t^2]i^ + [2.00m/s+(0.550m/s^2)t]j^, we substitute t = 7.93 s into the equation to find the x and y components of velocity:

Vx = 5.00m/s − (0.0180m/s^3)(7.93s)^2 = -3.84 m/s

Vy = 2.00m/s + (0.550m/s^2)(7.93s) = 6.91 m/s

Finally, we can find the magnitude of the velocity using the equation:

|V| = sqrt(Vx^2 + Vy^2) = sqrt((-3.84 m/s)^2 + (6.91 m/s)^2) = 8.05 m/s

B) What is the direction (in degrees counterclockwise from +x-axis) of the velocity of the car at t = 7.93 s?

To find the direction of the velocity, we can use the inverse tangent function:

θ = atan(Vy / Vx)

θ = atan(6.91 m/s / -3.84 m/s) = -60.6 degrees counterclockwise from the +x-axis.

C) What is the magnitude of the acceleration of the car at t = 7.93 s?

To find the magnitude of the acceleration, we need to calculate the magnitude of the acceleration at t = 7.93 s. The acceleration is given by the derivative of the velocity:

a =[dV/dt]i^ + [dV/dt]j^

a =[-(0.0360m/s^3)t]i^ + [(0.550m/s^2)]j^

Substituting t = 7.93 s into the equation, we find:

ax = -(0.0360m/s^3)(7.93s) = -0.285 m/s^2

ay = (0.550m/s^2)

Finally, we can find the magnitude of the acceleration using the equation:

|a| = sqrt(ax^2 + ay^2) = sqrt((-0.285 m/s^2)^2 + (0.550 m/s^2)^2) = 0.616 m/s^2

D) What is the direction (in degrees counterclockwise from +x-axis) of the acceleration of the car at t = 7.93 s?

To find the direction of the acceleration, we can use the inverse tangent function:

θ = atan(ay / ax)

θ = atan(0.550 m/s^2 / -0.285 m/s^2) = -63.7 degrees counterclockwise from the +x-axis.

Answer 2

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 = √(3.87^2 + 6.36^2)`

`v = 7.45 m/s`

Part b)

the direction of the velocity is given as

`\theta = tan^(-1)(6.36)/(3.87)`

`\theta = 58.7 degree`

part c)

for acceleration we know that

`a = (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 = √(0.285^2 + 0.550^2)`

`a = 0.62 m/s^2`

Part d)

for the direction of the motion

`\theta = tan^(-1)(0.550)/(-0.285)`

`\theta = 117.4 degree`