The position of a squirrel running in a park is given by r =[(0.280m/s) t +(0.03

ID: 1656805 • Letter: T

Question

The position of a squirrel running in a park is given by r =[(0.280m/s)t+(0.0360m/s^2)t^2]i^+ (0.0190m/s^3)t^3j^.

Part A

What is x(t), the x-component of the velocity of the squirrel, as function of time?

Part B

What is y(t), the y-component of the velocity of the squirrel, as function of time?

Part C

At 5.59 s , how far is the squirrel from its initial position?

Part D

At 5.59 s , what is the magnitude of the squirrel's velocity?

Part E

At 5.59 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

The position of a squirrel running in a park is given by r =[(0.280m/s)t+(0.0360m/s^2)t^2]i^+ (0.0190m/s^3)t^3j^.

Part A

What is x(t), the x-component of the velocity of the squirrel, as function of time?

Part B

What is y(t), the y-component of the velocity of the squirrel, as function of time?

Part C

At 5.59 s , how far is the squirrel from its initial position?

Part D

At 5.59 s , what is the magnitude of the squirrel's velocity?

Part E

At 5.59 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

r =[(0.280m/s)t+(0.0360m/s^2)t^2]i^+ (0.0190m/s^3)t^3j^.

a) dr/dt , since we need only the x component we will diffrentiateonly the i direction component

Vx= 0.280m/s)+(0.0360m/s^2)(2t)

b) similarily we diffrentiate only the j direction coponent of positionto get the velocity in y direction

Vy= (0.0190m/s^3)(3t^2)

c) put t= 5.59 in the position and distance at t =5.59 s from origin is = sqrt ( square of i component + square of j component )

d) tan theta = vy/vx =(0.0190m/s^3)(3t^2) / 0.280m/s)+(0.0360m/s^2)(2t)

where t = 5.59 s

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