Answer: 12.99 m/s
Explanation:
Since the acceleration is constant, we can use the following formula for the velocity
[tex]\vec{v}(t) = \vec{a}t + \vec{v}_0[/tex] (1)
Where:
[tex]\vec{v}(t) = \text{velocity at time }t
\newline \indent \vec{a} = \text{acceleration at time }t
\newline \indent \vec{v}_0 = \text{initial velocity, i.e., velocity at }t=0
[/tex]
Since at t = 10 s,
[tex]\vec{v}(10) = 33\textbf{i} + 15\textbf{j}
\newline \indent \vec{a} = 2.3\textbf{i} + 3.6\textbf{j}
[/tex]
the unknown variable in equation (1) is [tex]\vec{v}_0[/tex].
At t = 10 s, equation 1 becomes:
[tex]\vec{v}(10) = 10\vec{a} + \vec{v}_0[/tex] (2)
By subtracting both sides of equation (2) by [tex]10\vec{a}[/tex], we have
[tex]\vec{v}_0 = \vec{v}(10) - 10\vec{a} \newline \vec{v}_0 = 33\textbf{i} + 15\textbf{j} - 10(2.3\textbf{i} + 3.6\textbf{j}) \newline \vec{v}_0 = 33\textbf{i} + 15\textbf{j} - (23\textbf{i} + 36\textbf{j}) \newline \boxed{\vec{v}_0 = 10\textbf{i} - 21\textbf{j}}[/tex]
Now, we let
v(10) = speed of the object at t = 10
v(0) = speed of the object at t = 0
Recall that speed is the magnitude of the velocity and so
[tex]v(10) = |\vec{v}(10)|
\newline \indent \indent = |33\textbf{i} + 15\textbf{j}|
\newline \indent \indent = \sqrt{33^2 + 15^2}
\newline \indent \boxed{v(10) \approx 36.25}
\newline
\newline \indent v(0) = |\vec{v}(0)|
\newline \indent \indent = |10\textbf{i} - 21\textbf{j}|
\newline \indent \indent = \sqrt{10^2 + (-21)^2}
\newline \indent \boxed{v(0) \approx 23.26}
[/tex]
Hence, the change of speed after 10 s is given by
[tex]v(10) - v(0) \approx 36.25 - 23.26
\newline \indent \boxed{v(10) - v(0) \approx 12.99}
[/tex]
