Question

A record of travel along a straight path is as follows:
1. Start from rest with constant acceleration of 2.67 m/s2 for 12.0 s.
2. Maintain a constant velocity for the next 1.85 min.
3. Apply a constant negative acceleration of −8.77 m/s2 for 3.65 s.
(a) What was the total displacement for the trip?

Answer 1

Easy. We basically need to break this question into 3 parts.
All we need are the four kinematic equations and some basic unit calculations.

First, we can find the displacement for the constant acceleration at the beginning:
\[d_1=v_{i}t_1+\frac{1}{2}a_1t_1^2\]\[d_1=(0\frac{m}{s})(12.0s)+\frac{1}{2}(2.67\frac{m}{s^2})(12.0s)^2\]\[d_1=192.24m\]
Then, we need to find the final velocity from Part 1 to give us the constant velocity mentioned in Part 2.
\[v_{f}=v_{i}+a_1t_1\]\[v_{f}=(0\frac{m}{s})+(2.67\frac{m}{s^2})(12.0s)\]\[v_{f}=32.04\frac{m}{s}\]Now we just need to multiply by the time to get displacement.
\[d_2=v_{f}t_2\]The measurement of time from Part 2 is in minutes so we add a factor of 60 for the unit conversion.
\[d_2=(32.04\frac{m}{s})(1.85\times60s)\]\[d_2=3556.44m\]
Finally, we need to use the same kinematic equation from Part 1 to calculate the final distance.
\[d_3=v_ft_3+\frac{1}{2}a_3t_3^2\]It's important to make sure we calculate the acceleration as negative. Vectors have a funny way of working themselves out when you keep the signs correct.
\[d_3=(32.04\frac{m}{s})(3.65s)+\frac{1}{2}(-8.77\frac{m}{s^2})(3.65s)^2\]\[d_3=116.946m+58.4191625m\]Now, I'm stopping here because for the first time in our calculations we have to perform addition. That means that we must take into account significant figures. For Part 1 and Part 2, we were only performing multiplication so we could ignore the sigfig until the end and compare to the original data. Now, however, we need to take it into account. We know that all the givens have three sigfigs, so the results of multiplication are all 3-digits. Let's round, then add, then round again.
\[d_3=117m+58.4m\]\[d_3=175.4m\]\[d_3=175m\]
Then we take the sum of the three distances. Sums are addition, so now we need to take into account the sigfigs of the other distances as well.
\[d_1=192m\]\[d_2=3560m\]\[\Sigma d=d_1+d_2+d_3\]\[\Sigma d=192m+3560m+175m\]\[\Sigma d=3927m\]There's no rounding to be done with sigfig addition rules, so that's your final answer.