Question

1. using data collected in an experiment, a student determined the speed of his object at each time interval. ignore any acceleration due to a change in direction.

time in sec	velocity in m/s
1	2
2	2
3	5
4	1

a) during which time interval did the object move at constant speed?
b) when was the object’s rate of acceleration the greatest?
c) what was the object’s rate of acceleration between 3 and 4 seconds?
d) what was his displacement between 3 and 4 sec?

Explanation:

Step1: Identify constant - speed interval

Constant speed means velocity does not change. From the table, velocity is 2 m/s from \(t = 1\) s to \(t = 2\) s.

Step2: Calculate acceleration for each interval

The formula for acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_f - v_i\) and \(\Delta t=t_f - t_i\).
For \(0 - 1\) s: \(a_1=\frac{2 - 0}{1-0}=2\) m/s².
For \(1 - 2\) s: \(a_2=\frac{2 - 2}{2 - 1}=0\) m/s².
For \(2 - 3\) s: \(a_3=\frac{5 - 2}{3 - 2}=3\) m/s².
For \(3 - 4\) s: \(a_4=\frac{1 - 5}{4 - 3}=- 4\) m/s². The magnitude of \(a_4\) is 4 m/s², which is the greatest.

Step3: Find acceleration between 3 and 4 seconds

Using \(a=\frac{\Delta v}{\Delta t}\), with \(v_i = 5\) m/s, \(v_f = 1\) m/s, \(t_i = 3\) s, \(t_f = 4\) s. \(a=\frac{1 - 5}{4 - 3}=-4\) m/s².

Step4: Calculate displacement between 3 and 4 seconds

Since the motion is one - dimensional, and we can assume the motion is along a straight line. The average velocity \(\bar{v}=\frac{v_i + v_f}{2}=\frac{5+1}{2}=3\) m/s. Displacement \(d=\bar{v}\times\Delta t\), with \(\Delta t = 1\) s. So \(d = 3\times1 = 3\) m.

Answer:

a) 1 - 2 seconds
b) 3 - 4 seconds
c) - 4 m/s²
d) 3 m