Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Starting from rest a lady bug runs along a straight line with constant acceleration. After time T1 , the lady bug suddenly reverses the direction of its acceleration but keeps the magnitude of the acceleration constant and the same as it was before.

How much time (in terms of T1) would it take the ladybug to get back to its starting point from the place where it turned around?

share|improve this question

3 Answers 3

up vote 5 down vote accepted

The following is a Physics-oriented, more or less formula-free solution. When the acceleration changes, the velocity still goes the wrong way. By symmetry, after a further time $T_1$, the ladybug is at rest, twice as far from its starting point as the distance from the starting point to the place it stepped on the brakes.

So now it has twice that distance to traverse. Distance travelled from rest, under constant acceleration, is proportional to the square of time. So the time required to get back to the starting point from the furthest point reached is $\sqrt{2}\,T_1$. Total time from the time the acceleration reversed is therefore $T_1+\sqrt{2}\,T_1$.

share|improve this answer

You can also see it graphically: just use T1 as you time unit, and the distance travelled in that time as distance unit. Then the time required to return to the origin is below the red line. It's clearly equal to 1 (T1 ; the time required to reverse direction) plus the same time to travel a distance of 2 units (yellow line), which is $\sqrt{2}$.

enter image description here

share|improve this answer

The displacement of the ladybug from its starting position after time $t_1$ has elapsed is $$s_1 = \frac 1 2 a t_1^2$$

At the end of time $t_1$ it has a velocity, $$v = at_1$$

To find the time to cover the same distance back means to solve the equation $$\frac 1 2 a t_1^2 = - (at_1)t +\frac{1}{2}at^2 $$

To get the solutions $$t = t_1 (1-\sqrt{2})$$ $$t = t_1 (1+\sqrt{2})$$

The physical content of the problem dictates which solution to choose. However, I will leave it at that as this is a math forum. (Hint: Should $t$ be greater than or less than $t_1$?)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.