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A particle, P; is moving along the x-axis. The velocity of particle P at time is given by Vp(t) sin(,”5) for 0 <n<t At time t = 0. particle P is at position x = 5.
A second particle, Q, also moves along the x-axis. The velocity of partiele at time t is given by ve(t) (t _ 1.8) 1.25′ for 0 < t < t. At time t = 0_ particle Q is at position x = 10.
Find the positions of particles P and Q at time t = [.
(b) Are particles P and Q moving toward each other or away from each other at time =[ ? Explain your reasoning:
Find the acceleration of particle Q at time =IIs the speed of particle Q increasing or decreasing at time t = [ Explain your reasoning:
Find the total distance traveled by particle P over the time interval 0 < t < T_

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01:22

Let $s(t)=t /\left(t^{2}+5\right)$ be the position function of a particle moving along a coordinate line. where $s$ is in meters and $t$ is in seconds. Use a graphing utility to generate the graphs of $s(t), v(t),$ and $a(t)$ for $t \geq 0,$ and use those graphs where needed.(a) Use the appropriate graph to make a rough estimate of the time at which the particle first reverses the direction of its motion; and then find the time exactly.(b) Find the exact position of the particle when it first reverses the direction of its motion.(c) Use the appropriate graphs to make a rough estimate of the time intervals on which the particle is speeding up and on which it is slowing down; and then find those time intervals exactly.

01:09

Let $s(t)=1 / e^{t}$ be the position function of a particle moving along a coordinate line. where $s$ is in meters and $t$ is in seconds. Use a graphing utility to generate the graphs of $s(t), v(t),$ and $a(t)$ for $t \geq 0,$ and use those graphs where needed.(a) Use the appropriate graph to make a rough estimate of the time at which the particle first reverses the direction of its motion; and then find the time exactly.(b) Find the exact position of the particle when it first reverses the direction of its motion.(c) Use the appropriate graphs to make a rough estimate of the time intervals on which the particle is speeding up and on which it is slowing down: and then find those time intervals exactly.

02:41

A particle’s position on the $x$ -axis is given by the function $x=\left(t^{2}-4 t+2\right) \mathrm{m}$ where $t$ is in $\mathrm{s}$ a. Make a position-versus-time graph for the interval 0 s $\leq$ $t \leq 5$ s. Do this by calculating and plotting $x$ every 0.5 s from $0 \mathrm{s}$ to $5 \mathrm{s}$, then drawing a smooth curve through the points.b. Determine the particle’s velocity at $t=1.0 \mathrm{s}$ by drawing the tangent line on your graph and measuring its slope.c. Determine the particle’s velocity at $t=1.0 \mathrm{s}$ by evaluating the derivative at that instant. Compare this to your result from part b.d. Are there any turning points in the particle’s motion? If so, at what position or positions?e. Where is the particle when $v_{x}=4.0 \mathrm{m} / \mathrm{s} ?$f. Draw a motion diagram for the particle.

10:24

A particle moves according to a law of motion $s = f(t), t \ge 0,$ where $t$ is measured in seconds and $s$ in feet.(a) Find the velocity at time $t.$(b) What is the velocity after I second? (c) When is the particle at rest?(d) When is the particle moving in the positive direction?(e) Find the total distance traveled during the first 6 seconds.(f) Draw a diagram like Figure 2 to illustrate the motion of the particle.(g) Find the acceleration at time $t$ and after 1 second.(h) Graph the position, velocity, and acceleration functions for $0 \le t \le 6.$(i) When is the particle speeding up? When is it slowing down?

$f(t) = \frac {9t}{t^2 + 9}$

Transcript

1, so here we have been given the velocity function first, that is the sign t raised to 1.5, which is equal to the d x by d g, so this d x equals to sine t raised to 1.5 d g. So if we need a position, you must integrate. It t raised to 1.5 d, t 40 to 1. Second, this is the time so here, first of all, if i assume t raised to 1.5 that should be equal to the theta or some unknown constant. Let’S say d: sorry, let’s say y, so this 1.5 and…