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The position of a squirrel running in a park is given by r⃗ =[(0.280m/s)t+(0.0360m/s2)t2]i^+ (0.0190m/s3)t3j^.

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

What is , the -component of the velocity of the squirrel, as function of time?

vx(t)=0.280m/s

vx(t)=(0.0720m/s2)t

vx(t)=(0.280m/s)t+(0.0720m/s2)t2

vx(t)=0.280m/s+(0.0720m/s2)t

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

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

vy(t)=(0.0570m/s2)t2

vy(t)=(0.0570m/s3)t

vy(t)=(0.0570m/s3)t2

vy(t)=(0.0570m/s3)t+(0.0720m/s2)t2

At 5.99 s , how far is the squirrel from its initial position?

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Express your answer to three significant figures and include the appropriate units.

At 5.99 s , what is the magnitude of the squirrel’s velocity?

Express your answer to three significant figures and include the appropriate units.

At 5.99 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel’s velocity?

Express your answer to three significant figures and include the appropriate units.

• The position of a squirrel running in a park is given by r⃗ =[(0.280m/s)t+(0.0360m/s2)t2]i^+ (0.0190m/s3)t3j^.

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

vx(t)=0.280m/s

vx(t)=(0.0720m/s2)t

vx(t)=(0.280m/s)t+(0.0720m/s2)t2

vx(t)=0.280m/s+(0.0720m/s2)t ◄ this one, the first derivative of the i^ part

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

vy(t)=(0.0570m/s2)t2 ◄ this one is close, but the units are wrong!

vy(t)=(0.0570m/s3)t

vy(t)=(0.0570m/s3)t2 ◄ this one, the first derivative of the j^ part

vy(t)=(0.0570m/s3)t+(0.0720m/s2)t2

(iii) At 5.99 s , how far is the squirrel from its initial position?

r(x) = 0.280m/s*5.99s + 0.0360m/s²*(5.99s)² = 2.97 m

r(y) = 0.0190m/s³ * (5.99s)³ = 4.08 m

|r| = √(rx² + ry²) = 5.05 m

(iv) At 5.99 s , what is the magnitude of the squirrel’s velocity?

vx = 0.280m/s + 0.0720m/s²*5.99s = 0.711 m/s

vy = 0.0570m/s³ * (5.99s)² = 2.05 m/s

|v| = √(vx² + vy²) = 2.17 m/s

(v) At 5.99 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel’s velocity?

Θ = arctan(vy / vx) = arctan(2.05 / 0.711) = 70.8º

Hope this helps!