Question

name jostin alvarez aguasvivos mr. kenny college physics kinematics exam 1&2 dimensions 1. the position, x, of an object moving along the x - axis is shown above as a function of time t. the acceleration of this object must be: (a) zero b. constant but not zero c. increasing d. decreasing 2. a car is traveling east at 30 m/s before slowing to a stop in 3 seconds. its acceleration is: (a) 10 m/s/s east (b) 10 m/s/s west (c) 3 m/s/s east 3 m/s/s west (d) zero (e) not enough info 3. an object is dropped from rest. how far has it fallen after 1 second? (a) 0 m (b) 1 m (c) 4.9m (d) 9.8 m (e) 100m 5. which of the following pairs of graphs shows the distance traveled versus time and the speed versus time for an object uniformly accelerated from rest?

Explanation:

Question 1

Step1: Recall the relationship between position - time graph and acceleration

The slope of a position - time graph gives the velocity. If the position - time graph is a straight line, the velocity is constant.
For a straight - line position - time graph \(x = x_0+vt\) (where \(x_0\) is the initial position and \(v\) is the constant velocity). The acceleration \(a=\frac{dv}{dt}\). Since \(v\) is constant, \(\frac{dv}{dt} = 0\)

Step1: Use the formula for acceleration \(a=\frac{v - u}{t}\)

The initial velocity \(u = 30\ m/s\) (east), the final velocity \(v=0\ m/s\), and the time \(t = 3\ s\)

$$a=\frac{0 - 30}{3}=- 10\ m/s^{2}$$

The negative sign indicates that the acceleration is in the opposite direction of the initial velocity (west)

Step1: Use the kinematic equation \(h=v_0t+\frac{1}{2}gt^{2}\)

Since the object is dropped from rest, \(v_0 = 0\ m/s\), and \(g = 9.8\ m/s^{2}\), \(t = 1\ s\)

$$h=\frac{1}{2}(9.8)(1)^{2}=4.9\ m$$

Answer:

A. zero

Question 2