Question

the graph shows the motion of a student walking at different speeds. students motion during which time interval does the student walk with a negative acceleration? 0 - 2 seconds 2 - 4 seconds 4 - 8 seconds 8 - 12 seconds

Explanation:

Step1: Recall the formula for acceleration

Acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_{final}-v_{initial}\) and \(\Delta t=t_{final} - t_{initial}\). Negative acceleration means \(v_{final}

Step2: Analyze each time - interval

• For \(0 - 2\) seconds: Velocity \(v = 1\ m/s\) (constant). Using \(a=\frac{\Delta v}{\Delta t}\), \(\Delta v=1 - 1=0\), so \(a = 0\)
• For \(2 - 4\) seconds: Velocity is increasing (from \(1\ m/s\) to a higher value). Using \(a=\frac{\Delta v}{\Delta t}\), \(\Delta v>0\), so \(a>0\)
• For \(4 - 8\) seconds: Velocity \(v = 1.7\ m/s\) (constant). Using \(a=\frac{\Delta v}{\Delta t}\), \(\Delta v=1.7 - 1.7 = 0\), so \(a = 0\)
• For \(8 - 12\) seconds: Velocity is decreasing (from \(1.7\ m/s\) to \(0\ m/s\)). Using \(a=\frac{\Delta v}{\Delta t}\), \(\Delta v=0 - 1.7<0\), so \(a<0\)

Answer:

\(8 - 12\) seconds