Gievn:
![r^(\prime)^(\prime)=\lbrack4.0\operatorname{cm}+(2.5\operatorname{cm}/s^2)t^2\rbrack\text{ i +(5.0cm/s)tj}]()
Part A.
Let's find the magnitude of the dot's average velocity between t = 0 and t = 2.0s.
At t = 0:
Substitute 0 for t
![\begin{gathered} r_1=\lbrack4.0\operatorname{cm}+(2.5\operatorname{cm})0^2\rbrack i+(5.0\operatorname{cm})(0)j \\ \\ r_1=4.0\operatorname{cm}i+0\operatorname{cm}j \end{gathered}]()
At t = 2:
Substitute 2 for t
![\begin{gathered} r_2=\lbrack4.0\operatorname{cm}+(2.5\operatorname{cm})2^2\rbrack i+(5.0\operatorname{cm})(2)j \\ \\ r_2=\lbrack4.0\operatorname{cm}+10.0\operatorname{cm}\rbrack i+10.0\operatorname{cm}j \\ \\ r_2=14.0\operatorname{cm}i+10.0\operatorname{cm}j \end{gathered}]()
To find the magnitude of the dot's average velocity, apply the formula:
![\begin{gathered} V_(ar)=|(r_2-r_1)/(t_2-t_1)| \\ \\ V_(ar)=|\frac{(14.0\operatorname{cm}i+10.0\operatorname{cm}j)-(4.0\operatorname{cm}i+0\operatorname{cm}j)}{2-0}| \\ \\ V_(ar)=|\frac{14.0\operatorname{cm}i-4.0\operatorname{cm}i+10.0\operatorname{cm}j-0\operatorname{cm}j}{2-0}| \\ _{} \\ \\ V_(ar)=|\frac{10.0\operatorname{cm}i+10.0\operatorname{cm}j}{2}| \\ \\ V_(ar)=|\frac{10.0\operatorname{cm}i}{2}+\frac{10.0\operatorname{cm}j}{2}| \\ \\ V_(ar)=|5.0\operatorname{cm}i+5.0\operatorname{cm}j| \\ \\ V_(ar)=\sqrt[]{5^2+5^2} \\ \\ V_(ar)=\sqrt[]{25+25} \\ \\ V_(ar)=\sqrt[]{50} \\ \\ V_(ar)=7.1\operatorname{cm}\text{ /s} \end{gathered}]()
Therefore, the magnitude of the dot's average velocity is approximately 7.1 cm/s.
Part B.
FInd the direction angle of the dot's average velocity between t = 0s and t = 2.0 s.
To find the direction angle between t = 0s and t = 2.0s, we have:
Take the inverse tangent of both sides:
Therefore, the direction of the dot's average velocity between t = 0 and t = 2.0 s is 45 degrees.
ANSWER:
• a) 7.1 cm/s
• b) 45 degrees.