Question

a car goes from 39 km/hr to a complete stop in 15 seconds. what is the acceleration?

Explanation:

Step1: Convert initial velocity to m/s

Initial velocity \( u = 39\space km/hr=\frac{39\times1000}{3600}\space m/s=\frac{390}{36}\space m/s=\frac{65}{6}\space m/s\approx10.833\space m/s\)
Final velocity \( v = 0\space m/s\) (since it stops)
Time \( t = 15\space s\)

Step2: Use acceleration formula \( a=\frac{v - u}{t}\)

\(a=\frac{0-\frac{65}{6}}{15}=\frac{-\frac{65}{6}}{15}=-\frac{65}{90}=-\frac{13}{18}\approx - 0.722\space m/s^{2}\)

Answer:

The acceleration is approximately \(-0.72\space m/s^{2}\) (or \(-\frac{13}{18}\space m/s^{2}\) or in km/hr²: first find change in velocity is \(- 39\space km/hr\) in \(15\space s=\frac{15}{3600}\space hr\), so \(a=\frac{-39}{\frac{15}{3600}}=-39\times\frac{3600}{15}=-9360\space km/hr^{2}\))