Initial velocity when deacelerating at a constant rate

[xc630]

Homework Statement

You are driving along a straight section of roadway (x-axis). You spot a police officer and apply the brakes, slowing down at a constant rate. Your positions (in meters) at successive time intervals of 0.435 s are tabulated below as function of time. (1) Calculate the initial speed (i.e., the speed at t=0, the time at which the brakes are first applied.). (2) Calculate your acceleration along the x-axis, ax.
x (m) 7.00 19.73 31.92 43.55
t (s) 0.000 0.435 0.870 1.305

Homework Equations

v= (d2-d1)/t
x= Vo(t) +1/2at^2
V= Vo + at

The Attempt at a Solution

For the initial velocity I tried using the equation x= Vo (t) +1/2 at^2
the problem says slowing down at a constant rate so I put a= 0, leaving x=vo (t) I subtracted 43.55-7 to get 36.55m over a period of the 1.305s. Dividing 36.55/1.305 i got 28 m/s but the correct one is supposed to be 29.9m/s. What did I do wrong. And for part B of the problem would I just be able to use the initial velocity from part A and the same equation I mentioned above to find a?

 

[Andrew Mason]

x (m) 7.00 19.73 31.92 43.55
t (s) 0.000 0.435 0.870 1.305

..

The Attempt at a Solution

For the initial velocity I tried using the equation x= Vo (t) +1/2 at^2
the problem says slowing down at a constant rate so I put a= 0,

Why would a = 0 if the car is slowing down? The rate of change of a may be 0 but not a.

AM

 

[tomcue]

Your approach to finding the initial velocity was correct. However, the value you got for the initial velocity (28 m/s) is not the same as the correct value (29.9 m/s) because you rounded the value of the time interval (1.305 s) to only three significant figures. This may seem like a small difference, but when dealing with calculations involving acceleration, even small differences in values can lead to significant differences in the final result. To get a more accurate value for the initial velocity, you should use the full value of the time interval (1.305 s) without rounding.

For part B of the problem, you can indeed use the initial velocity you calculated in part A and the same equation (x= Vo (t) +1/2 at^2) to find the acceleration. Rearranging the equation to solve for acceleration, we get a=(2x-Vo(t))/t^2. Plugging in the values from part A (x=36.55 m, Vo=29.9 m/s, t=1.305 s), we get a= -2.96 m/s^2. This means that your acceleration along the x-axis is -2.96 m/s^2, which indicates that you are decelerating at a constant rate of 2.96 m/s^2.