Question

name: ______________
test # 2b
the following four questions refer to the motion of a go - cart. the graph shown below is a plot of the car’s velocity as a function of time.
graph: velocity (m/s) vs time (s), with grid, time from 0.0 to 7.0 s, velocity from - 4.0 to 4.0 m/s

1. at which time is the go - karts acceleration zero?

a) 2.0 s b) 3.0 s c) 4.0 s d) 6.0 s e. none of these

1. what is the go - karts displacement over the first 5.0 s?

a) - 6.0m b) - 4.0m c) - 2.0m d) 4.0m e) 6.0m

1. what distance does the go - karts cover in the first 7.0 s (assume it starts from 0 m)?

a) - 5.0 m b) 2.0 m c) 8.5 m d) 17 m e) 22 m

1. what is the go - karts instantaneous acceleration at 6.0 s?

a) - 2.0 m/s² b) - 1.0 m/s² c) 0 m² d) 2.0 m/s² e) 4.0 m/s²

1. a baseball is thrown vertically upward into the air. what is the instantaneous acceleration of the ball at its highest point?

a) zero
b) 9.8 m/s2 up
c) 9.8 m/s2 down
d) changing from 9.8 m/s² up to 9.8 m/s² down

Explanation:

Question 12

Step1: Recall acceleration definition

Acceleration is the slope of the velocity - time graph. When acceleration is zero, the slope of the velocity - time graph is zero (the graph is horizontal).

Step2: Analyze each option

• Option A: At \(t = 2.0\space s\), the velocity - time graph is changing (slope is non - zero), so acceleration is non - zero.
• Option B: At \(t=3.0\space s\), the velocity - time graph is horizontal (slope \( = 0\)), so acceleration \(a = 0\space m/s^{2}\).
• Option C: At \(t = 4.0\space s\), the velocity - time graph is sloping (non - zero slope), so acceleration is non - zero.
• Option D: At \(t=6.0\space s\), the velocity - time graph is sloping (non - zero slope), so acceleration is non - zero.

Step1: Recall displacement from v - t graph

Displacement is the area under the velocity - time graph. We can divide the graph from \(t = 0\) to \(t = 5.0\space s\) into geometric shapes (rectangles and triangles or trapezoids).

• From \(t = 0\) to \(t=1.5\space s\) (approx, looking at the graph, the first horizontal segment: velocity \(v_1=1.0\space m/s\) (since from \(y\) - axis, the first horizontal line is at \(v = 1.0\space m/s\)) and time \(t_1 = 1.5\space s\) (from \(0\) to \(1.5\space s\) approximately, since at \(t = 1.5\space s\) (mid - point of \(0 - 3\space s\) grid? Wait, the grid: each small square on x - axis: from \(0\) to \(1.0\space s\) is 5 small squares, so each small square is \(0.2\space s\). On y - axis, from \(0\) to \(2.0\space m/s\) is 5 small squares, so each small square is \(0.4\space m/s\). Wait, maybe better to calculate the area as sum of positive and negative areas.
• From \(t = 0\) to \(t = 1.5\space s\) (let's say the first horizontal part: from \(t = 0\) to \(t = 1.5\space s\), velocity \(v = 1.0\space m/s\) (since the first horizontal line is at \(y = 1.0\space m/s\) (between \(0\) and \(2.0\space m/s\), mid - point? Wait the graph: first horizontal segment: from \(t = 0\) to \(t = 1.5\space s\) (approx \(1.5\space s\)) with \(v = 1.0\space m/s\). Then from \(t = 1.5\space s\) to \(t = 2.5\space s\), it's a line from \(v = 1.0\space m/s\) to \(v=- 1.0\space m/s\) (slope), then from \(t = 2.5\space s\) to \(t = 3.5\space s\), horizontal at \(v=-1.0\space m/s\), then from \(t = 3.5\space s\) to \(t = 4.5\space s\), slope to \(v=-2.0\space m/s\), then from \(t = 4.5\space s\) to \(t = 5.5\space s\), horizontal at \(v = - 2.0\space m/s\) (but we need up to \(t = 5.0\space s\)).

Wait, a better approach: The area under the v - t graph is displacement. Let's calculate the area in each interval:

1. Interval 1: \(t = 0\) to \(t = 1.5\space s\) (let's say the first horizontal part, from the graph, the first horizontal line is from \(t = 0\) to \(t = 1.5\space s\) (since at \(t = 1.5\space s\) (3 small squares on x - axis, each \(0.5\space s\)? Wait the x - axis: from \(0\) to \(1.0\space s\) is 5 small squares, so each square is \(0.2\space s\). So from \(0\) to \(1.5\space s\) is \(1.5\space s\) (7.5 small squares? No, maybe the first horizontal segment is from \(t = 0\) to \(t = 1.5\space s\) (since at \(t = 1.5\space s\) (3 units of \(0.5\space s\)) with \(v = 1.0\space m/s\). Area \(A_1=v\times t=1.0\times1.5 = 1.5\space m\) (positive area).

1. Interval 2: \(t = 1.5\space s\) to \(t = 2.5\space s\): This is a triangle (or trapezoid) with base \(t = 1.0\space s\) and height from \(v = 1.0\space m/s\) to \(v=-1.0\space m/s\), so the average velocity is \(\frac{1.0+( - 1.0)}{2}=0\), area \(A_2 = 0\space m\).

1. Interval 3: \(t = 2.5\space s\) to \(t = 3.5\space s\): Horizontal line at \(v=-1.0\space m/s\), time \(t = 1.0\space s\), area \(A_3=-1.0\times1.0=- 1.0\space m\) (negative area).

1. Interval 4: \(t = 3.5\space s\) to \(t = 4.5\space s\): A line from \(v=-1.0\space m/s\) to \(v = - 2.0\space m/s\), average velocity \(\frac{-1.0+( - 2.0)}{2}=-1.5\space m/s\), time \(t = 1.0\space s\), area \(A_4=-1.5\times1.0=-1.5\space m\).

1. Interval 5: \(t = 4.5\space s\) to \(t = 5.0\space s\): Horizontal line at \(v=-2.0\space m/s\), time \(t = 0.5\space s\), area \(A_5=-2.0\times0.5=-1.0\space m\).

Now sum all areas: \(A = A_1+A_2+A_3+A_4+A_5=1.5 + 0-1.0-1.5 - 1.0=-2.0\space m\).

Wait, maybe a simpler way: The graph from \(0\) to \(5\space s\):

• From \(0\) to \(1.5\space s\): rectangle with \(v =…

Step1: Recall distance from v - t graph

Distance is the sum of the absolute values of the areas under the velocity - time graph (since distance is a scalar, we consider the magnitude of the area for each segment).

We divide the time from \(t = 0\) to \(t = 7.0\space s\) into segments:

1. Segment 1: \(t = 0\) to \(t = 1.5\space s\): Velocity \(v = 1.0\space m/s\), time \(t = 1.5\space s\), area (distance part) \(A_1=|1.0\times1.5| = 1.5\space m\).

1. Segment 2: \(t = 1.5\space s\) to \(t = 2.5\space s\): The graph goes from \(v = 1.0\space m/s\) to \(v=-1.0\space m/s\). The area (distance) is the area of the triangle formed, with base \(t = 1.0\space s\) and height \(h = 2.0\space m/s\) (from \(y = 1\) to \(y=-1\)). Area \(A_2=\frac{1}{2}\times1.0\times2.0 = 1.0\space m\).

1. Segment 3: \(t = 2.5\space s\) to \(t = 3.5\space s\): Velocity \(v=-1.0\space m/s\), time \(t = 1.0\space s\), area \(A_3=|-1.0\times1.0| = 1.0\space m\).

1. Segment 4: \(t = 3.5\space s\) to \(t = 4.5\space s\): The graph goes from \(v=-1.0\space m/s\) to \(v=-2.0\space m/s\). Area (distance) is the area of the trapezoid, with \(v_1=-1\), \(v_2=-2\), \(t = 1.0\space s\). Area \(A_4=\frac{|-1|+|-2|}{2}\times1.0=\frac{1 + 2}{2}\times1=1.5\space m\).

1. Segment 5: \(t = 4.5\space s\) to \(t = 5.5\space s\): Velocity \(v=-2.0\space m/s\), time \(t = 1.0\space s\), area \(A_5=|-2.0\times1.0| = 2.0\space m\).

1. Segment 6: \(t = 5.5\space s\) to \(t = 6.0\space s\): The graph goes from \(v=-2.0\space m/s\) to \(v = 0\space m/s\) (wait, no, at \(t = 6.0\space s\) it crosses zero and goes to positive. Wait, from \(t = 5.5\space s\) to \(t = 6.0\space s\), velocity goes from \(-2.0\space m/s\) to \(0\space m/s\). Area (distance) is the area of the triangle with base \(t = 0.5\space s\) and height \(h = 2.0\space m/s\). Area \(A_6=\frac{1}{2}\times0.5\times2.0 = 0.5\space m\).

1. Segment 7: \(t = 6.0\space s\) to \(t = 7.0\space s\): Velocity \(v = 2.0\space m/s\), time \(t = 1.0\space s\), area \(A_7=|2.0\times1.0| = 2.0\space m\). Wait, no, from \(t = 6.0\space s\) to \(t = 7.0\space s\), the velocity is \(2.0\space m/s\) (horizontal line). Wait, maybe my earlier segmentation is wrong. Let's use a better grid:

Each small square on x - axis: \(0.2\space s\) (since from \(0\) to \(1.0\space s\) is 5 squares: \(1.0\div5 = 0.2\space s\) per square).

Each small square on y - axis: \(0.4\space m/s\) (since from \(0\) to \(2.0\space m/s\) is 5 squares: \(2.0\div5 = 0.4\space m/s\) per square).

Now, let's calculate the area for each part:

• From \(t = 0\) to \(t = 1.5\space s\) (7.5 squares on x - axis): Velocity is \(1.0\space m/s\) (2.5 squares on y - axis: \(2.5\times0.4 = 1.0\space m/s\)). Area (distance) \(A_1=1.0\times1.5 = 1.5\space m\) (positive area, magnitude).

• From \(t = 1.5\space s\) to \(t = 2.5\space s\) (5 squares on x - axis, \(t = 1.0\space s\)): Velocity goes from \(1.0\space m/s\) to \(-1.0\space m/s\) (change of \(2.0\space m/s\) over \(1.0\space s\)). The area (distance) is the area of the triangle: \(\frac{1}{2}\times1.0\times2.0 = 1.0\space m\) (since it's a triangle with base \(1.0\space s\) and height \(2.0\space m/s\)).

• From \(t = 2.5\space s\) to \(t = 3.5\space s\) (5 squares on x - axis, \(t = 1.0\space s\)): Velocity is \(-1.0\space m/s\) (2.5 squares on y - axis, magnitude \(1.0\space m/s\)). Area \(A_3=1.0\times1.0 = 1.0\space m\).

• From \(t = 3.5\space s\) to \(t = 4.5\space s\) (5 squares on x - axis, \(t = 1.0\space s\)): Velocity goes from \(-1.0\space m/s\) to \(-2.0\space m/s\) (change of \(1.0\space m/s\) over \(1.0…

Answer:

B. 3.0 s

Question 13