Can anyone finally help me with this Calculus proof?

Question

Using derivatives a = dv/dt, v = ds/dt and |v| = speed, show that if an object is moving on a straight line, then its speed is increasing when its velocity v and its acceleration a have the same sign and its speed is decreasing when v and a have opposite sign.

 

I’m taking the first class in Calculus and this problem is from applications of antidifferentiation. My teacher says that no fancy Trigonometry is necessary. There is a hint: If |v| = S, when v > 0, S = v and dS/dt = dv/dt = a. When v < 0, S = - v and dS/dt = - dv/dt = - a...but I'm not seeing how to do this.

 

 

Answer 1

Hi Jaina,

 

This has more to do with your intuitive-level understanding of vector than it does about calculus. 

1) Drop a stone from a height h and you know that it's speed increases as it approaches the ground. What can we deduce from this? from this fact that the speed increases as it approaches the ground? We know that the velocity and the acceleration are both negative (conventionally up is positive and dawn is negative). Therefore, you have them both with same sign (negative in this case) you have the speed decreasing.

2) What about as the stone going up? Remember, acceleration (due to gravity) does not change. It is still negative. But, the velocity (because it's going up) is positive. What can we ay about the speed as it goes up before it stops? Decreasing, right. Therefore, opposite sign and decreasing speed (negative acceleration and positive velocity).  I hope this helps you