Question

Joshua saw his girl from the distance. He was initially walking at 2 m/s westward. However, when he saw his true love, his final speed reached 5 m/s. If he accelerated at 1.8 m/s, how long did it take him to reach his girlfriend?

Answer 1

Answer:

Initial velocity = 2 m/s

Final velocity = 5 m/s

Acceleration = 1.8 $m/s^{2}$

Therefore, time taken = $\frac{v - u}{a} = \frac{5 - 2}{1.8} = \frac{3}{1.8} OR \frac{30}{18}$

= 1.6 (bar over 6) seconds