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 xcomponent 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 ycomponent of the velocity of the squirrel, as function of time?
What is , the ycomponent 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?
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 +xaxis) of the squirrel’s velocity?
Express your answer to three significant figures and include the appropriate units.
1 Answer

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 xcomponent 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 ycomponent 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 +xaxis) of the squirrel’s velocity?
Θ = arctan(vy / vx) = arctan(2.05 / 0.711) = 70.8º
Hope this helps!