Instantaneous Velocity and Speed Problems and Solution

 Problem#1

An electron moving along the x axis has a position given by x = 16te^–t m, where t is in seconds. How far is the electron from the origin when it momentarily stops?

Answer:
a position given by x = 16te^–t m, then velocity is
v = dx/dt = 16e^-t + (-1)(16t)e^-1
v = 16e^-1(1 – t)

the time it takes for the electron to stop is (when v = 0)
0 = 16e^-1(1 – t)
t = 1 s

so, the distance that electrons take to stop is

x = 16e^–1  = 5.9 m

Problem#2
(a) If a particle’s position is given by x = 4 – 12t + 3t² (where t is in seconds and x is in meters), what is its velocity at t = 1 s? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time t; if not, answer no. (f) Is there a time after t = 3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer no.

Answer:
(a) The velocity of the particle is
v = dx/dt = d(4 – 12t + 3t²)/dt = –12 + 6t
Thus, at t = 1 s, the velocity is v = (–12 + (6)(1)) = –6 m/s.

(b) Since v < 0, it is moving in the –x direction at t = 1 s.

(c) At t = 1 s, the speed is |v| = 6 m/s.

(d) For 0 < t < 2 s, |v| decreases until it vanishes. For 2 < t < 3 s, |v| increases from zero to the value it had in part (c). Then, |v| is larger than that value for t > 3 s.

(e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to positive (note that as t → + ∞, we have v → + ∞). One can check that v = 0 when t = 2 s.

(f) No. In fact, from v = –12 + 6t, we know that v > 0 for t > 2 s.

Problem#3
The position function x(t) of a particle moving along an x axis is x = 4.0 – 6.0t² , with x in meters and t in seconds. (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin? (e) Graph x versus t for the range -5 s to +5 s. (f) To shift the curve rightward on the graph, should we include the term +20t or the term -20t in x(t)? (g) Does that inclusion increase or decrease the value of x at which the particle momentarily stops?

Answer:
We use the functional notation x(t), v(t), and a(t) in this solution, where the latter two quantities are obtained by differentiation:

v = dx/dt = d(4.0 – 6.0t²)/dt = –12t

and
a = dv/dt = d(–12t)/dt = –12 m/s²

(a) From v(t) = 0 we find it is (momentarily) at rest at t = 0.

(b) We obtain x(0) = 4.0 m

(c) and (d) Requiring x(t) = 0 in the expression x(t) = 4.0 – 6.0t 2 leads to t = ±0.82 s for the times when the particle can be found passing through the origin.

(e) We show both the asked-for graph (on the left) as well as the “shifted” graph (Fig.1) that is relevant to part (f). In both cases, the time axis is given by –3 ≤ t ≤ 3 (SI units understood).

(f) We arrived at the graph on the right (shown above) by adding 20t to the x(t) expression.

(g) Examining where the slopes of the graphs become zero, it is clear that the shift causes the v = 0 point to correspond to a larger value of x (the top of the second curve shown in part (e) is higher than that of the first).

Fig.1

Problem#4
The position of a particle moving along the x axis is given in centimeters by x = 9.75 + 1.50t³ , where t is in seconds. Calculate (a) the average velocity during the time interval t = 2.00 s to t = 3.00 s; (b) the instantaneous velocity at t = 2.00 s; (c) the instantaneous velocity at t = 3.00 s; (d) the instantaneous velocity at t = 2.50 s; and (e) the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s. (f) Graph x versus t and indicate your answers graphically.

Answer:
(a) We plug into the given equation for x for t = 2.00 s and t = 3.00 s and obtain x₂ = 21.75 cm and x₃ = 50.25 cm, respectively. The average velocity during the time interval 2.00 ≤ t ≤ 3.00 s is
v_avg = ∆x/∆t = (50.25 cm – 21.75 cm)/(3.00 s – 2.00 s) = 28.5 cm/s

(b) The instantaneous velocity is
 v = dx/dt = = 4.5t² , which, at time t = 2.00 s, yields v = (4.5)(2.00)² = 18.0 cm/s.

(c) At t = 3.00 s, the instantaneous velocity is v = (4.5)(3.00)² = 40.5 cm/s

(d) At t = 2.50 s, the instantaneous velocity is v = (4.5)(2.50)² = 28.1 cm/s.

(e) Let tm stand for the moment when the particle is midway between x₂ and x₃ (that is, when the particle is at x_m = (x₂ + x₃)/2 = 36 cm). Therefore,
x_m = 9.75 + 1.5t_m³
36 cm = 9.75 + 1.5t_m³
1.5t_m³ = 26.25
t_m = 2.596 s

Thus, the instantaneous speed at this time is v = 4.5(2.596)² = 30.3 cm/s.

Fig.2

(f) The answer to part (a) is given by the slope of the straight line between t = 2 and t = 3 in this x-vs-t plot. The answers to parts (b), (c), (d), and (e) correspond to the slopes of tangent lines (not shown but easily imagined) to the curve at the appropriate points.