Question

the figure shows the y-position (in blue) of a particle versus time. (a) what is the average velocity of the particle during the time interval t = 1.50 s to t = 4.50 s? (express your answer in vector form.) (b) using the tangent to the curve (shown as the orange line in the figure), what is the instantaneous velocity of the particle at t = 1.50 s? (express your... (c) at what time is the velocity of the particle equal to zero?

Explanation:

Part (a)

Step1: Determine the y - positions at \( t_1 = 1.50\ s \) and \( t_2 = 4.50\ s \)

From the graph, we can estimate the y - position. Let's assume that at \( t = 1.50\ s \), the y - position \( y_1 \) is \( 8\ m \) (by looking at the grid, at \( t = 1.5\ s \), the blue curve and the orange line intersect around \( y = 8\ m \)) and at \( t = 4.50\ s \), the y - position \( y_2 \) is \( 18\ m \) (from the graph, at \( t = 4.5\ s \), the blue curve is around \( y = 18\ m \)). The time interval \( \Delta t=t_2 - t_1=4.50 - 1.50 = 3.00\ s \). The displacement \( \Delta y=y_2 - y_1=18 - 8 = 10\ m \).

Step2: Calculate average velocity

The formula for average velocity \( \vec{v}_{avg}=\frac{\Delta\vec{y}}{\Delta t} \). Since the motion is along the y - axis, the average velocity vector is in the y - direction. So \( \vec{v}_{avg}=\frac{10\ m}{3.00\ s}\hat{j}\approx3.33\hat{j}\ m/s \) (if we consider more accurate reading: Let's re - check the graph. The orange line (tangent) has a slope. Let's find the y - values more accurately. At \( t = 0 \), the orange line starts at \( y = 2\ m \) (approx) and at \( t = 3\ s \), it is at \( y = 14\ m \) (approx). Wait, maybe a better way: The average velocity is \( \frac{y_2 - y_1}{t_2 - t_1} \). Let's take from the graph, at \( t = 1.5\ s \), the blue curve is at \( y_1 = 8\ m \) (since at \( t = 1\ s \), it's at \( 6\ m \), \( t = 2\ s \) at \( 12\ m \), so linear approximation gives \( y(1.5)=8\ m \)). At \( t = 4.5\ s \), the blue curve is at \( y_2 = 18\ m \) (at \( t = 4\ s \), it's at \( 17\ m \), \( t = 5\ s \) at \( 15\ m \), so maybe \( 18\ m \) at \( t = 4.5\ s \)). Then \( \Delta y=18 - 8 = 10\ m \), \( \Delta t = 3\ s \), so \( v_{avg}=\frac{10}{3}\approx3.33\ m/s \) in y - direction, so \( \vec{v}_{avg}=3.33\hat{j}\ m/s \) (or more accurately, if we take the orange line's slope for average? Wait, no, average velocity is displacement over time. Let's do it properly. Let's find two points on the blue curve: at \( t_1 = 1.5\ s \), \( y_1 \): looking at the grid, each square is \( 2\ m \) in y and \( 0.5\ s \) in t. At \( t = 1.5\ s \) (which is \( 3 \) units of \( 0.5\ s \)), the y - position is \( 8\ m \) (4 units of \( 2\ m \)). At \( t = 4.5\ s \) (9 units of \( 0.5\ s \)), the y - position is \( 18\ m \) (9 units of \( 2\ m \)). So \( \Delta y=18 - 8 = 10\ m \), \( \Delta t=4.5 - 1.5 = 3\ s \). So \( v_{avg}=\frac{10}{3}\approx3.33\ m/s \), so in vector form \( \vec{v}_{avg}=3.33\hat{j}\ m/s \) (or \( \frac{10}{3}\hat{j}\approx3.33\hat{j}\)).

Part (b)

Step1: Determine the slope of the tangent line (orange line)

The orange line (tangent) passes through two points. Let's take \( t = 0\ s \), \( y = 2\ m \) and \( t = 3\ s \), \( y = 14\ m \) (from the graph). The slope of the line (which is the instantaneous velocity at \( t = 1.5\ s \) since it's the tangent) is given by \( v=\frac{\Delta y}{\Delta t} \).

Step2: Calculate the slope

\( \Delta y = 14 - 2=12\ m \), \( \Delta t=3 - 0 = 3\ s \). So \( v=\frac{12}{3}=4\ m/s \). Since it's along the y - axis, the instantaneous velocity is \( \vec{v}=4\hat{j}\ m/s \).

Part (c)

Step1: Recall that velocity is the slope of the y - t graph

The velocity of the particle is zero when the slope of the \( y - t \) curve is zero (horizontal tangent). From the graph, the blue curve has a maximum (horizontal tangent) at \( t = 4.0\ s \) (looking at the graph, the peak of the curve is at \( t = 4\ s \), where the slope is zero).

Part (a) Answer: \( \boldsymbol{3.33\hat{j}\ (or\ \frac{10}{3}\hat{j})}\)

Part (b) Answer: \( \boldsymbol{4\hat{j}}\)

Part (c) Answer: \( \boldsymbol{4.0}\) (approx, depending on graph reading, could be around \( 4.0\ s \))

Answer:

Step1: Recall that velocity is the slope of the y - t graph

The velocity of the particle is zero when the slope of the \( y - t \) curve is zero (horizontal tangent). From the graph, the blue curve has a maximum (horizontal tangent) at \( t = 4.0\ s \) (looking at the graph, the peak of the curve is at \( t = 4\ s \), where the slope is zero).

Part (a) Answer: \( \boldsymbol{3.33\hat{j}\ (or\ \frac{10}{3}\hat{j})}\)

Part (b) Answer: \( \boldsymbol{4\hat{j}}\)

Part (c) Answer: \( \boldsymbol{4.0}\) (approx, depending on graph reading, could be around \( 4.0\ s \))