A cat walks in a straight line, which we shall call thex-axis with the positive direction to the right. As an observantphysicist, you make measurements of this cat's motion and con-struct a graph of the feline's velocity as a function of time(Fig. E2.30). (a) Find the cat's velocity at t = 4.0 s and att = 7.0 s. (b) What is the cat's acceleration at t = 3.0 s? Att = 6.0 s? At t = 7.0 s? (c) What distance does the cat move dur-ing the first 4.5 s? From t = 0 tot = 7.5 s? (d) Sketch clear graphsof the cat's acceleration and position as functions of time, assumingthat the cat started at the origin.Figure E2.30U₂(cm/s)878765432-OTT-T12341567t (s)

The slope of velocity time graph is called acceleration. To find the values of velocity at any…

Answered: A cat walks in a straight line, which…

Similar questions

Nerve impulses in a human body travel at a speed of about 100 m/s. Suppose a woman accidentally steps barefoot on a thumbtack. About how much time does it take the nerve impulse to travel from the foot to the brain (in s)? Assume the woman is 1.80 m tall and the nerve impulse travels at uniform speed.
In heavy rush-hour traffic, you drive in a straight line at 12 m/s for 1.5 minutes, then you have to stop for 3.5 minutes, and finally, you drive at 15 m/s for another 2.5 minutes. Plot a position-versus-time graph for this motion. Your plot should extend from t = 0 minutes to t = 7.5 minutes. Assume x=0 and t=0 at the start of your motion.
A jeep travels a distance d=22.1m in the positive x direction in a time t1=20.2s, at which point the jeep brakes, coming to rest in t2=7.38s. 1. What was the jeep's instantaneous velocity in the horizontal direction, in meters per second, when it began braking? 2. Using the result from question 1, what was the jeep's horizontal component of acceleration, in meters per squared second, during the braking period?
A cannonball is fired at 171 (m/s), straight up in the air, by accident. Quickly realizing the error, the commanding officer orders the gunners to fire another cannonball straight up to hit the one coming down. In order to be safe, the two cannon balls must explode at a given hight of 636 (m). What is the maximum time (in seconds) the second cannonball can be fired after the first, such that the two cannaballs collide, exploding safely above the regiment? (s) How fast (velocity) will the first cannon ball be traveling just before the collision? (m/s) How fast (velocity) will the second cannon ball be traveling just before the collision? (m/s)
Suppose you are standing at the edge of a 200 meter tall cliff and are in the habit of dropping objects from great heights. So, you drop a rock off of the cliff. (a) What are the initial position s(0), velocity v(0), and acceleration a(0) of the rock? b) Use your answers from part (a) to define functions s(t), v(t), and a(t) describing the position, velocity, and acceleration, respectively, after t seconds. (c) How long will it take for the rock to hit the ground? How fast is it falling when it hits?
Hello. I am working on a problem with motion. The questions asks me to calculate the maximum height (h1), total time (t2), and speed of a ball right before it hits the ground. The question states that A person is throwing a ball upward into the air with an initial speed Vo = 10m/s. Assume that the instant when the ball is released, the person's hand is at a height ho = 1.5m. The speed of the ball at its peak height is zero, and the question needs to be solved in ascending part and descending part. I don't understand how to solve for the maximum height. What is the correct formula to use and why? For other questions like this, I will be able to solve them if I know the formulas for the ascending of the ball and the descent of the ball as well as the explanation. Thank you. For the sake of the question, the ball is being thrown straight up.
An object accelerating at constant acceleration (a) has an initial velocity v0 (at t=0), a velocity v at time t, and goes a particular distance d or x - x0 during the time interval. Derive the expression relating v0, v, a, and d (or x-x0). The relation must not include time. You may use either algebra or calculus.
A particle moves along the x-axis according to the equation x 2.00 + 3.00t - the acceleration of the particle. 1.00t2. where x is in meters and t is in seconds. At t=D3.00s, find
A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 2.8 seconds the rocket is at a height of 91 meters. (A) What are the magnitude and direction of the rocket's acceleration? (B) What is it's speed at this time? Can you please solve the question using one or more of these equations: v = v0 +at vav = (1/2) times (v0 + v) x = x0 + (1/2) times (v0 + v) times t x = x0 + v0t + (1/2)at2 v2 = v02 + 2a(x - x0)
An object's position as a function of time in one dimension is given by the expression; 2.74t2 + 3.01t + 13.9 where are constants have proper SI Units. What is the object's average velocity between the times t = 4.06 s and t = 8.22 s?
An object's position in the X-direction as a function of time is given by the expressio%3B x(t) = 5t2 + 2t where are quantities have proper SI Units. What is the object's average velocity in the x-direction between the times t = 1.71 s and t = 2.37 s. Just enter the number rounded to 3 significant figures and assume it has proper SI Units.
The acceleration of a particle is a constant. At t = 0 the velocity of the particle is (15.8î + 18.4ĵ) m/s.At = 3.6 s the velocity is 10.5ĵ m/s. (Use the following as necessary: t. Do not include units in your answers.) (b) How do the position (in m) and velocity (in m/s) vary with time? Assume the particle is initially at the origin.

SEE MORE QUESTIONS