Thread: body starting from rest

1. body starting from rest

A body starting from rest travels in a straight line, first with uniform acceleration a and then with uniform deceleration b.
It comes to rest when it has covered a total distance d.

If the overall time for the journey is T, show that t^2 =2d (1/a +1/b)

i have the solution to the question but cant understand it as the person who did them doesnt have clear writing and also didnt explain everything he did here is the link.
http://thephysicsteacher.ie/Exam%20M...ation/1979.pdf its the second question on the link can somebody explain how he writes distance under the curve=2d . it says the total distance travelled was d so why did he say 2d. also how did he come up with t1=(b/b+a)T

2. Re: body starting from rest

You need to pay more attention to upper and lower case variables ...

What is to be shown is $T^2=2d\left(\dfrac{1}{a}+\dfrac{1}{b}\right)$

for accelerations $a$ and $b$ ...

$v_{max}=0+ at_1 \text{ and } 0= v_{max} - bt_2 \implies at_1 = bt_2 \implies t_2 = \dfrac{at_1}{b}$

$T = t_1+t_2 = t_1 + \dfrac{at_1}{b} = \dfrac{bt_1}{b} + \dfrac{at_1}{b} = t_1 \cdot \dfrac{b+a}{b} \implies t_1 = \dfrac{b}{b+a} \cdot T$

distance, $d$, is the area of the triangular region under the velocity graph ...

$d=\dfrac{1}{2} \cdot T \cdot v_{max}$

$d=\dfrac{1}{2} \cdot T \cdot (0 + at_1)$

$2d = T \cdot at_1$

$2d = T \cdot a \cdot \dfrac{b}{b+a} \cdot T$

$2d = T^2 \cdot \dfrac{ab}{b+a}$

$T^2 = 2d \cdot \dfrac{b+a}{ab}$

$T^2 = 2d \left(\dfrac{b}{ab} + \dfrac{a}{ab}\right)$

$T^2 = 2d \left(\dfrac{1}{a} + \dfrac{1}{b}\right)$

3. Re: body starting from rest

I understand it but would you say that's the best way to go about the question or is there an easier way. It seems like a round about way to do the question

4. Re: body starting from rest

To get $T^2$ in terms of distance, $d$, and the magnitudes of the accelerations, $a$ and $b$, I would say it's an efficient method. The best? ... I can't say until someone shows a better way.