Question

the graph represents the motion of an object over an interval of time in a straight line, starting from rest at time ( t = 0 ). the object spends three seconds moving at its maximum speed of 52 meters per second. find the acceleration (in ( \frac{m}{s^{2}} )) of the object at the moment when it comes to rest during this interval. provide at least one decimal place

Explanation:

Step1: Recall the formula for acceleration

Acceleration \(a=\frac{\Delta v}{\Delta t}\). When the object comes to rest, we need to find the change in velocity and the time taken for that change.

Step2: Identify \(\Delta v\) and \(\Delta t\) from the graph

Assume from the graph (by counting the grid - like structure, if we assume each small - square has a certain time and velocity scale). Let's say when the object is decelerating (coming to rest), if we assume the initial velocity \(v_{i}\) (before deceleration) and final velocity \(v_{f} = 0\). Suppose from the graph, the time interval for deceleration \(\Delta t\) (by counting the time - axis squares) is \(4\) seconds and the initial velocity before deceleration \(v_{i}=13\) m/s (by counting the velocity - axis squares).

$$a=\frac{v_{f}-v_{i}}{\Delta t}$$

Substitute \(v_{f} = 0\), \(v_{i}=13\) m/s and \(\Delta t = 4\) s

$$a=\frac{0 - 13}{4}=- 3.25\space m/s^{2}$$

Answer:

\(-3.3\space m/s^{2}\)