PHY2053 midterm (ch. 1-5)
Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed up, but then she hears over the PA system that her filght is delayed. The acceleration versus time graph shows Emmy's acceleration during the 5 s5 s time interval in which all of the events occur. In the velocity versus time graph, construct a plot of Emmy's velocity over this same time period. Assuming all the values given are exact, what is Emmy's velocity at a time of 3.59 s? Enter your answer to at least three significant digits.
Acceleration describes the rate at which velocity changes with time. Therefore, the area under the acceleration versus time graph equals the change in velocity. If the area is positive (above the time axis), then the velocity increases; if the area is negative (below the time axis), then the velocity decreases. For example, between 00 and 1 s1 s , the area under the acceleration versus time graph is a positive 2 m/s.2 m/s. Therefore, Emmy's velocity increases by 2 m/s2 m/s between 00 and 1 s1 s . Between 11 and 2 s2 s , the area under the acceleration versus time graph is 1 m/s1 m/s , so the velocity increases again during this time period, but only by 1 m/s1 m/s . The acceleration is zero from 2 s2 s to 3 s3 s , so the velocity does not change during this time period. The area under the graph between 33 and 4 s4 s is negative 3 m/s3 m/s , so the velocity during this time decreases by 3 m/s3 m/s . Finally, between 44 and 5 s5 s , the acceleration is again zero, so Emmy's velocity does not change during this final time period. Note that the acceleration is constant over each second‑long time period, so the velocity during each period changes at a constant rate. To find Emmy's velocity at time 𝑡=3.59 st=3.59 s , use the velocity versus time graph to see that her speed at 3 s3 s is 5 m/s5 m/s . Between 33 and 4 s4 s , her velocity 𝑣3-4v3-4 decreases by 3 m/s3 m/s each second (the acceleration is −3 m/s2−3 m/s2 during this time period), so her velocity between 33 and 4 s4 s is given by 𝑣3-4(𝑡)=(5 m/s)−(3 m/s2)×(𝑡−3 s)v3-4(t)=(5 m/s)−(3 m/s2)×(t−3 s) At 3.59 s3.59 s , her velocity is 𝑣3-4(at 𝑡=3.59 s)=(5 m/s)−(3 m/s2)×(3.59 s−3 s)=3.23 m/s
Consider the 𝑥-𝑡x-t plot shown. For each defined interval identify the sign (that is, + or −+ or −) of velocity 𝑣v and acceleration 𝑎.a. Between 0 s0 s and 2 s,2 s, the velocity is _______ and the acceleration is _________ Between 4 s4 s and 5 s,5 s, the velocity is _________ and the acceleration is __________ For times greater than 5.8 s,5.8 s, the velocity is ___________ and the acceleration is ___________
An object's velocity corresponds to the slope of the tangent line on an x-t, or position versus time, graph. When the slope of the tangent line is positive, that is, when the final position is greater than the initial position during a given time interval, the object's velocity is positive. Conversely, when the slope is negative, the object's velocity is negative. In the x-t plot shown, the slope of the tangent line, and therefore the velocity, is negative from 0 s to 2 s. The slope and the velocity are positive during the next time interval, from 4 s to 5 s, and are negative for the last time interval, beyond 5.8 s, as the line approaches zero. Acceleration is the change in velocity per unit time. Since the slope of the tangent line on an x-t graph represents velocity, the change in the slope represents acceleration. Therefore, on a position versus time graph, the change in the slope is represented by the curvature of the curve. Positive acceleration on an x-t graph is indicated by a concave upward curve because velocity changes in the positive direction, whereas negative acceleration on an x-t graph is indicated by a concave downward curve because velocity changes in the negative direction. In the x-t plot shown, the acceleration is negative between 0 s and 2 s, since the line curves downward and the negative slope becomes even more negative. Between 4 s and 5 s, the acceleration is negative as the positive slope decreases to zero. For times greater than 5.8 s, the acceleration is positive as the line curves upward and the negative slope increases to zero.
You notice a one-dimensional-thinking snail crawling along one rail of a railroad track. Naturally, you move it to a safer place. But before you do, you observe it as it moves from position 𝑥i=−92.1 cm to position 𝑥f=−65.7 cm along the rail, as measured from a nearby joint between two segments of rail. What is the magnitude of the snail's displacement, Δ𝑥, in centimeters?
For the given initial and final positions 𝑥ixi and 𝑥fxf, respectively, the displacement Δ𝑥Δx is Δ𝑥=𝑥f−𝑥iΔx=xf−xi Enter the data to obtain the numerical answer. Δ𝑥=(−65.7 cm)−(−92.1 cm)=26.4 cm
The figure shows a position versus time graph for a red blood cell leaving the heart. Determine the instantaneous speed 𝑣v of the red blood cell when 𝑡=13 ms.
The instantaneous speed 𝑣v of an object describes how its position 𝑥x changes with respect to time 𝑡t at a single instant in time. It is different from average speed 𝑣average,vaverage, which describes the constant speed required to change the position of an object by an amount Δ𝑥Δx over some time interval Δ𝑡.Δt. 𝑣average=Δ𝑥Δ𝑡vaverage=ΔxΔt To relate instantaneous speed to average speed, consider a time interval Δ𝑡Δt centered around a time 𝑡=𝑡0t=t0 on a graph of position versus time. Generally speaking, the average speed 𝑣averagevaverage over the interval Δ𝑡Δt does not equal the instantaneous speed at 𝑡0,t0, 𝑣(𝑡0).v(t0). However, the smaller the size of the interval Δ𝑡Δt about 𝑡0,t0, the closer 𝑣averagevaverage is to 𝑣(𝑡0).v(t0). For very small values of Δ𝑡,Δt, 𝑣average≈𝑣(𝑡0),vaverage≈v(t0), and the slope 𝑚m of the position function at 𝑡0t0 is approximately equal to 𝑣average.vaverage. 𝑣average=Δ𝑥Δ𝑡≈𝑚(𝑡0) for very small Δ𝑡vaverage=ΔxΔt≈m(t0) for very small Δt Therefore, the instantaneous speed 𝑣(𝑡0)v(t0) equals the slope of the position function at 𝑡0.t0. To determine the instantaneous speed of a red blood cell at 𝑡=13 ms,t=13 ms, first draw a line tangent to the position curve of the red blood cell at time 𝑡=13 ms.t=13 ms. Next, determine two points on the line tangent to the position curve of the red blood cell. The points on the tangent line in the figure are (𝑡1,𝑥1)(𝑡2,𝑥2)=(8.0 ms,1.0 mm)=(13.0 ms,6.5 mm)(t1,x1)=(8.0 ms,1.0 mm)(t2,x2)=(13.0 ms,6.5 mm) Substitute the values for the coordinates into the equation for the slope and solve. 𝑚=6.5 mm−1.0 mm13.0 ms−8.0 ms≈1.0 m/sm=6.5 mm−1.0 mm13.0 ms−8.0 ms≈1.0 m/s Note that 1 mm/ms1 mm/ms equals 1 m/s.1 m/s. As mentioned previously, the instantaneous speed is the absolute value of the slope. 𝑣=|𝑚|≈1.0 m/sv=|m|≈1.0 m/s Therefore, the instantaneous speed of the red blood cell at 𝑡=13 mst=13 ms is about 1.0 m/s.
Position is 0 for 5 seconds. John is rollerblading down a long, straight path. At time zero, there is a mailbox about 1 m1 m in front of him. In the 5 s5 s time period that follows, John's velocity is given by the velocity versus time graph in the figure. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph. Assuming that all the numbers given are exact, what is John's position at a time of 4.37 s? Enter your answer to at least three significant digits.
Velocity is the rate at which position changes with time. The area under the velocity versus time graph during some time interval is equal to the amount by which the position changes in that time interval. Areas above the time axis are positive, and correspond to positive changes in position; areas below the time axis are negative, and correspond to negative changes in position. For example, between times 𝑡=0t=0 and 𝑡=1,t=1, there are two boxes between the velocity curve and the time axis, therefore the area enclosed by the velocity versus time graph is −2 m,−2 m, meaning John's position decreases by 2 m.2 m. Since John starts at −1 m−1 m at time 𝑡=0t=0, his position one second later is −3 m.−3 m. John is at rest between 𝑡=1t=1 and 𝑡=2t=2 seconds, so his position remains at −3 m−3 m during that interval. After 2 s,2 s, John's velocity is 2 m/s,2 m/s, so his position increases by 2 m2 m each second thereafter. John's position at 2 s2 s is −3 m,−3 m, and after 2 s2 s his velocity is 2 m/s,2 m/s, so his position at 4.37 s4.37 s is 𝑥(at 𝑡=4.37 s)=(−3 m)+(2ms)(4.37 s−2 s)=1.74 m
