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.

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: let's assume the first horizontal part is from \(t = 0\) to \(t = 1.5\space s\) with \(v=1.0\space m/s\) (since from the grid, each square is, say, \(0.5\space s\) on x - axis and \(0.5\space m/s\) on y - axis). Wait, more accurately, let's look at the graph:
• From \(t = 0\) to \(t = 1.5\space s\) (let's say the first horizontal line: velocity \(v_1=1.0\space m/s\) (since between \(0\) and \(2\) s, the velocity is positive, around \(1\space m/s\)), time \(t_1 = 1.5\space s\) (approx, but maybe better to count the squares. Let's assume each small square on x - axis is \(0.25\space s\) and on y - axis is \(0.5\space m/s\).
• From \(t = 1.5\space s\) to \(t = 2.5\space s\): velocity decreases from \(1\space m/s\) to \(- 1\space m/s\) (a triangle? No, a trapezoid? Wait, maybe a better approach:
• The area above the time axis (positive velocity) and below (negative velocity).
• From \(t = 0\) to \(t = 1.5\space s\): rectangle with \(v = 1\space m/s\), \(t=1.5\space s\), area \(A_1=1\times1.5 = 1.5\space m\)
• From \(t = 1.5\space s\) to \(t = 2.5\space s\): triangle with base \(1\space s\), height \(1\space m/s\) (going from \(1\) to \(0\)) and then a triangle going from \(0\) to \(- 1\space m/s\) in \(0.5\space s\)? Wait, maybe the correct way is:
• Let's split the time from \(0\) to \(5\space s\) into intervals:
• Interval 1: \(t = 0\) to \(t = 1.5\space s\), \(v = 1\space m/s\) (horizontal line). Area \(A_1=1\times1.5 = 1.5\space m\)
• Interval 2: \(t = 1.5\space s\) to \(t = 2.5\space s\), velocity goes from \(1\space m/s\) to \(- 1\space m/s\). This is a triangle with base \(1\space s\) and height \(1\space m/s\) (but since it's going from positive to negative, the area above the axis and below: the area above is a triangle with base \(0.5\space s\) (from \(1.5\) to \(2.0\)) and height \(1\space m/s\), area \(A_{2a}=\frac{1}{2}\times0.5\times1 = 0.25\space m\). The area below is a triangle with base \(0.5\space s\) (from \(2.0\) to \(2.5\)) and height \(1\space m/s\), area \(A_{2b}=\frac{1}{2}\times0.5\times1=0.25\space m\) (but negative)
• Interval 3: \(t = 2.5\space s\) to \(t = 3.5\space s\), \(v=-1\space m/s\) (horizontal line). Area \(A_3=-1\times1=- 1\space m\)
• Interval 4: \(t = 3.5\space s\) to \(t = 4.5\space s\), velocity goes from \(- 1\space m/s\) to \(- 2\space m/s\). Triangle with base \(1\space s\) and height \(1\space m/s\) (negative area), area \(A_4=\frac{1}{2}\times1\times(- 1)=- 0.5\space m\)
• Interval 5: \(t = 4.5\space s\) to \(t = 5.0\space s\), \(v = - 2\space m/s\) (horizontal line). Time \(t = 0.5\space s\), area \(A_5=-2\times0.5=-1\space m\)

• Now sum all areas: \(A_1+A_{2a}+A_{2b}+A_3+A_4+A_5=1.5 + 0.25-0.25-1 - 0.5-1=-1\space m\)? Wait, maybe my square counting is wrong. Let's use a better method. The velocity - time graph:

• From \(0\) to \(1.5\) s: velocity \(v = 1\space m/s\) (let's say each big square on x is \(1\) s, y is \(2\) m/s? Wait the graph has y - axis from - 4 to 4, x from 0 to 7. Let's assume each small square is \(0.5\) s (x - axis) and \(0.5\) m/s (y - axis).

• From \(t = 0\) to \(t = 1.5\) s (3 small squares on x - axis, since \(0.5\times3 = 1.5\) s), velocity \(v = 1\space m/s\) (2 small squares on y - axis, \(0.5\time…

Step1: Recall distance from v - t graph

Distance is the total area between the velocity - time graph and the time axis (regardless of sign, we take the absolute value of each area segment and sum them).

Step2: Divide the time from \(t = 0\) to \(t = 7.0\space s\) into intervals:

• Interval 1: \(t = 0\) to \(t = 1.5\space s\): velocity \(v = 1\space m/s\) (positive), area \(A_1=1\times1.5 = 1.5\space m\)
• Interval 2: \(t = 1.5\) to \(t = 2.5\space s\): velocity changes from \(1\) to \(- 1\) m/s. The area above the axis (positive

Answer:

B. 3.0 s

Question 13