Question

16 a car is traveling east. the graph shows the change in the velocity of the car over time, relative to the ground. graph: velocity (m/s) vs time (s) with a line from (0,0) to (6, 30) approximately during this period, what is the eastward acceleration of the car relative to the ground? f 5 m/s² g 25 m/s² h 30 m/s² j 150 m/s²

Explanation:

Step1: Recall the formula for acceleration

Acceleration \( a \) is defined as the change in velocity \( \Delta v \) divided by the change in time \( \Delta t \), i.e., \( a=\frac{\Delta v}{\Delta t} \).

Step2: Identify values from the graph

From the velocity - time graph, at \( t = 0\) s, the velocity \( v_0=0\) m/s. At \( t = 5\) s, the velocity \( v = 25\) m/s (we can also use \( t = 6\) s, but let's use \( t = 5\) s for simplicity). So, \( \Delta v=v - v_0=25 - 0 = 25\) m/s and \( \Delta t = 5-0 = 5\) s.

Step3: Calculate acceleration

Using the formula \( a=\frac{\Delta v}{\Delta t} \), substitute \( \Delta v = 25\) m/s and \( \Delta t=5\) s. Then \( a=\frac{25}{5}=5\) m/s². We can also check with another point, for example, at \( t = 2\) s, \( v = 10\) m/s. Then \( \Delta v=10 - 0 = 10\) m/s, \( \Delta t = 2\) s, and \( a=\frac{10}{2}=5\) m/s².

Answer:

F \( 5\space m/s^{2} \)