(a) [tex]t=\frac{v}{a_1}+\frac{v}{a_2}[/tex]
In the first part of the motion, the car accelerates at rate [tex]a_1[/tex], so the final velocity after a time t is:
[tex]v = u +a_1t[/tex]
Since it starts from rest,
u = 0
So the previous equation is
[tex]v= a_1 t[/tex]
So the time taken for this part of the motion is
[tex]t_1=\frac{v}{a_1}[/tex] (1)
In the second part of the motion, the car decelerates at rate [tex]a_2[/tex], until it reaches a final velocity of v2 = 0. The equation for the velocity is now
[tex]v_2 = v - a_2 t[/tex]
where v is the final velocity of the first part of the motion.
Re-arranging the equation,
[tex]t_2=\frac{v}{a_2}[/tex] (2)
So the total time taken for the trip is
[tex]t=\frac{v}{a_1}+\frac{v}{a_2}[/tex]
(b) [tex]d=\frac{v^2}{2a_1}+\frac{v^2}{2a_2}[/tex]
In the first part of the motion, the distance travelled by the car is
[tex]d_1 = u t_1 + \frac{1}{2}a_1 t_1^2[/tex]
Substituting u = 0 and [tex]t_1=\frac{v}{a_1}[/tex] (1), we find
[tex]d_1 = \frac{1}{2}a_1 \frac{v^2}{a_1^2} = \frac{v^2}{2a_1}[/tex]
In the second part of the motion, the distance travelled is
[tex]d_2 = v t_2 - \frac{1}{2}a_2 t_2^2[/tex]
Substituting [tex]t_2=\frac{v}{a_2}[/tex] (2), we find
[tex]d_1 = \frac{v^2}{a_2} - \frac{1}{2} \frac{v^2}{a_2} = \frac{v^2}{2a_2}[/tex]
So the total distance travelled is
[tex]d= d_1 +d_2 = \frac{v^2}{2a_1}+\frac{v^2}{2a_2}[/tex]
