# Kinematics of a particle in a straight line

A car travelling on a straight road slows with constant deceleration. The car passes a road sign with a speed of $\frac{100}{9} \ m/s$ and a post box with speed of $\frac{20}{3} \ m/s$. The distance between the road sign and the post box is $240\ m$. Find, in $ms^{-2}$, the deceleration of the car.

For this problem, I sketched a graph of velocity against time. I made time = 0 be when the velocity is $\frac{100}{9} \ m/s$. I found the time at the post box to be 27 secs, which is definitely correct. I then proceeded to find the slope of this linear velocity graph, since this would give the constant acceleration. $$\frac{\frac{100}{9} - \frac{20}{3}}{0-27} = -0.164... \ m/s^2$$

Now, I keep being told in problems like this that the acceleration would the positive value of what I got, but everything disagrees with this! My sketch shows a linear graph with a downward slope, implying the slope is negative. So, why is the answer $0.164... \ m/s^2$

The problem required you to find the deceleration which in the context of this question was meant to be the negative acceleration by the problem setter and so a positive answer was required.

Have a look at @dmckee 's comments as to why the term deceleration should not be used.

A small point is that your equation from finding the acceleration in that it is probably better that it should reflect the convention that when a change in something is being considered it is usual to have the change equal to the final value minus the initial value not the other way round.

$$\frac{\frac{20}{3}-\frac{100}{9}}{27-0} = -0.164... \rm m/s^2$$

This obviously makes no difference to the answer that you obtained.

Acceleration is normally defined as a vector quantity. It has a magnitude and a direction. The magnitude of acceleration is always a nonnegative number, but it has a direction as well (in one dimension, the direction corresponds to its sign).

You can perhaps better understand it by using the 3rd equation of motion??

Since velocity, acceleration and displacement all are vector quantities , you must first assign a positive or negative direction, taking in consideration the question. Lets assume that the direction from the road sign to the post box is the $+ve$ direction. In this case we can write the 3rd equation as :

$v_f = +\frac{20}{3}m/s$

$v_i = +\frac{100}{9}m/s$

$\Delta s = Displacement = +240m$

Note that the + sign indicates that the velocity/displacement is in the direction we have assumed to be +ve.(from the road sign to the post box)

Now applying and solving :-

$v_f^2 = v_i^2 + 2a\Delta s$

we get

$a = -0.164m/s^2$

But here the sign is -ve. This means that the magnitude of acceleration is $0.164$, but it's direction is negative. Since we had assumed the direction from the road sign to the post box as +ve, the negative sign applies that the acceleration is in opposite direction, i.e. opposite to the velocity of the car.

If you want, you can reverse the direction and end up getting the acceleration with a positive sign. However, in this case also, the acceleration would be opposite to the velocity of the car, which direction we had assumed to be negative.

Your sketch is absolutely correct, as since the acceleration is in the opposite direction, the slope of the v-t graph which gives acceleration, would be negative(since in this problem, the acceleration is in the opposite direction of the motion of the car)