# Using a vt graph to find the value of a velocity V after a time T with an acceleration of a, and starting velocity at 10 m/s

I am trying to model a situation in which a car passes the point A on a straight road with a speed of 10 m/s, and moves with a constant acceleration of a m/s^2 along the road for T seconds until it reaches the point B where its speed is V m/s. The car is then travelling at this speed (V) for a further 10 seconds when it reaches a point C. From C it travels for a further T seconds with a constant acceleration 3a m/s^2, until a speed of 20 m/s is reached at a point D. The question being asked is to show that V = 12.5 m/s. However, I feel as if I am doing the graph wrong. Here is what my graph looks like:

Is this graph ok, and if so what do I need to do in order to find V, because I have been trying to use three SUVAT equations as there are three variables (which means I need three equation to solve simultaneously) to get V, a and T, however, I haven't really made any progress. Is there a way to find a solution using the graph?

• Yes the question is to find V – Benny Jul 8 '18 at 4:30
• Th distance between A and D is given in the second part of the question, it is 675m – Benny Jul 8 '18 at 4:35

$$\triangle V =at+3at=20-10=10\mbox{ m/s}$$ $$\mbox{So, }at=2.5 \mbox{ m/s}$$
$$V=10+at=12.5\mbox{ m/s}$$ $$675=10t+\dfrac12at^2+(12.5)(10)+12.5t+\left(\dfrac12\right)3at^2$$ Now plug in $at=2.5$ $$675=10t+\frac12(2.5t)+125+12.5t+\frac32(2.5t)$$ Now solve for $t$
• $\triangle V =at+3at$ – Key Flex Jul 8 '18 at 5:16