Another Look at Constant Acceleration — Question 1

Question 1

Integration constants belong to the chosen clock

A point on an east-positive line has constant acceleration $a=+2\ \mathrm{m/s^2}$. At clock reading $t_0=4\ \mathrm{s}$ it has $v=-3\ \mathrm{m/s}$ and $x=10\ \mathrm{m}$. The model holds throughout $4\leq t\leq7\ \mathrm{s}$.

Acceleration and velocity satisfy $dv/dt=a$ and $dx/dt=v$. An antiderivative reverses differentiation: if $F'=f$, then $\int f(t)\,dt=F(t)+C$. You may use $\int t^n\,dt=t^{n+1}/(n+1)+C$ for nonnegative integers $n$, and integrate sums term by term.

1. Integrate twice with two independent constants. Determine them from the observations at 4 s, writing explicit expressions for $v(t)$ and $x(t)$ in the original clock.

2. Repeat using elapsed time $\tau=t-4\ \mathrm{s}$. Explain why the integration constants in the original clock need not equal the stated velocity and position at 4 s.

3. Find position and velocity at 7 s. Check the two initial observations and the required acceleration by differentiation and substitution.

4. A student writes $v=at-3\ \mathrm{m/s}$ and $x=\tfrac12at^2-(3\ \mathrm{m/s})t+10\ \mathrm{m}$. Identify precisely the wrong initial event built into these expressions.

Question 1 - Solution

Integration constants belong to the chosen clock

(a) Each integration introduces its own constant.

Using numerical seconds for $t$ and the corresponding SI units for the functions, integration gives $$v(t)=2t+C_1,\qquad x(t)=t^2+C_1t+C_2.$$ At 4 s, $-3=8+C_1$, so $C_1=-11$. Then $10=16-44+C_2$, so $C_2=38$. Thus $$\boxed{v(t)=(2t-11)\ \mathrm{m/s}},\qquad \boxed{x(t)=(t^2-11t+38)\ \mathrm{m}}.$$ Here $C_1$ is a velocity constant and $C_2$ a position constant; their units differ.

(b) Start the elapsed-time clock at the observation event.

With dimensional $\tau=t-4\ \mathrm{s}$, $$\boxed{v=-3\ \mathrm{m/s}+(2\ \mathrm{m/s^2})\tau},$$ $$\boxed{x=10\ \mathrm{m}-(3\ \mathrm{m/s})\tau+(1\ \mathrm{m/s^2})\tau^2}.$$ The constants now equal the observed values because $\tau=0$ at that event. In the original clock, the constant terms refer algebraically to $t=0$, not to 4 s; no physical extension to that earlier time is needed to impose the data on the stated interval.

(c) Verify both the data and the differential equations.

$$x'(t)=2t-11=v(t),\qquad v'(t)=2=a,\quad(x(4),v(4))=(10,-3)\quad\text{(SI)}.$$

At 7 s, $\boxed{v=+3\ \mathrm{m/s},\ x=10\ \mathrm{m}}$. At 4 s the expressions give $v=-3\ \mathrm{m/s}$ and $x=10\ \mathrm{m}$. Differentiating $x$ gives $v$, and differentiating $v$ gives the required constant $+2\ \mathrm{m/s^2}$.

(d) The student’s constants refer to a different event.

Those expressions set $v(0)=-3\ \mathrm{m/s}$ and $x(0)=10\ \mathrm{m}$. They do not impose these values at 4 s: they instead give $v(4)=5\ \mathrm{m/s}$ and $x(4)=14\ \mathrm{m}$. Integration constants must be fitted to the actual time attached to each initial condition.