Question

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Answer 1

a. Total Distance covered from AB = \(300\) m 
Total time taken = \(2 \times 60 + 30\) s 
=\(150\) s 

Therefore, Average Speed from AB = \(\frac{Total Distance }{ Total Time}\) 
=\(\frac{300 }{ 150}\) \(m s ^{-1}\) 
=\(2\) \(m s^{-1}\) 
Therefore, Velocity from AB =\(\frac{Displacement \;AB }{ Time}\) = \(\frac{300 }{ 150}\) \(m s^{-1}\) 
=\(2 m s^{-1}\) 
Total Distance covered from AC =AB + BC 
= \(300 + 200\) m 
Total time taken from A to C = Time taken for AB + Time taken for BC 
= \((2 \times 60+30)+60\) \(s\) 
= \(210\) \(s\) 
Therefore, Average Speed from AC = \(\frac{Total\, Distance }{Total \,Time}\) 
=\(\frac{ 400 }{210}\) \(m s^{-1}\) 
= \(1.904\) \(m s^{-1}\) 

---

b. Displacement (S) from A to C = AB - BC 
= \(300-100\) m = \(200\) m 
Time (t) taken for displacement from AC = \(210\) \(s\) 
Therefore, Velocity from AC = \(\frac{Displacement (s) }{ Time(t)}\) 
= \(\frac{200 }{ 210}\) \(m s^{-1}\) 
= \(0.952\) \(m s^{-1}\)