9C - Displacement, velocity and acceleration

9C - Displacement, velocity and acceleration

Can a moving particle have zero acceleration, or pass through rest without changing direction? You will use differentiation and integration to connect position, velocity and acceleration, then distinguish the distance travelled from the displacement between two times.

Velocity and acceleration

The sign of a velocity tells you the direction of motion along a line, while acceleration tells you how that velocity is changing.

Choose an origin OO and one positive direction. The coordinate s(t)s(t) is the particle's displacement from the origin at time tt. It may be positive or negative. The displacement from time aa to time bb is s(b)−s(a)s(b)-s(a).

The derivative links

v=dsdt,a=dvdt=d2sdt2.v=\frac{\mathrm{d}s}{\mathrm{d}t},\qquad a=\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{\mathrm{d}^2s}{\mathrm{d}t^2}.

With ss in metres and tt in seconds, vv is in m s−1\mathrm{m\,s^{-1}} and aa in m s−2\mathrm{m\,s^{-2}}. Speed is ∣v∣|v|, so it is nonnegative.

Positive acceleration means that velocity is increasing: for example, changing from −5-5 to −3-3 is an increase in velocity but a decrease in speed. When velocity and acceleration have opposite signs, the particle is slowing down. When they have the same sign, it is speeding up. At an instant of rest, v=0v=0; acceleration need not be zero.

Worked example 1: moving with zero acceleration

Question. A particle has s=t3−6t2+9t+2s=t^3-6t^2+9t+2, in metres, for t≥0t\ge0 seconds. Find its velocity, speed and acceleration at t=2t=2, and state its direction of motion.

Plan. Differentiate once for velocity and again for acceleration. Use the sign of velocity for direction.

v=3t2−12t+9,a=6t−12.v=3t^2-12t+9,\qquad a=6t-12.

At t=2t=2,

v=12−24+9=−3,a=12−12=0.v=12-24+9=-3,\qquad a=12-12=0.

Answer. Velocity −3 m s−1-3\,\mathrm{m\,s^{-1}}, speed 3 m s−13\,\mathrm{m\,s^{-1}}, acceleration 0 m s−20\,\mathrm{m\,s^{-2}}. The particle is moving in the negative direction.

Check. s(2+h)=4−3h+h3s(2+h)=4-3h+h^3, so the average velocity over this short interval is −3+h2-3+h^2, tending to −3-3 as hh tends to zero. Zero acceleration here does not mean zero velocity.

Now keep negative velocity but make acceleration positive. This distinguishes direction of motion from the direction in which velocity changes.

Worked example 2: negative velocity and positive acceleration

Question. A particle has s=12−5t+t2s=12-5t+t^2, in metres, for t≥0t\ge0 seconds. Find its velocity and acceleration at t=1t=1. Is it speeding up or slowing down there?

Plan. Differentiate and compare the signs of vv and aa.

v=−5+2t,a=2.v=-5+2t,\qquad a=2.

At t=1t=1, v=−3v=-3 and a=2a=2. Acceleration increases velocity towards zero, reducing its magnitude.

Answer. v=−3 m s−1v=-3\,\mathrm{m\,s^{-1}} and a=2 m s−2a=2\,\mathrm{m\,s^{-2}}. The particle is moving in the negative direction and slowing down.

Check. At t=0.9t=0.9, speed is 3.2 m s−13.2\,\mathrm{m\,s^{-1}}; at t=1.1t=1.1, speed is 2.8 m s−12.8\,\mathrm{m\,s^{-1}}. These nearby speeds confirm the decrease.

Apply the same sign distinction to this new displacement function.

Recovering motion from rates

Integration reverses each derivative link, but an initial condition is needed to locate the particular motion.

Integrating acceleration gives velocity plus a constant. Integrating velocity gives displacement from the origin plus another constant. Use the initial velocity to determine the first and the initial position to determine the second. They are separate constants, even if the same letter is sometimes used in different calculations.

Motion from an integral

v=∫a dt,s=∫v dt.v=\int a\,\mathrm{d}t,\qquad s=\int v\,\mathrm{d}t.

Include a constant in each indefinite integral. For a time interval [a,b][a,b],

s(b)−s(a)=∫abv(t) dt.s(b)-s(a)=\int_a^b v(t)\,\mathrm{d}t.

The definite integral gives a displacement during the interval. Add the starting position to obtain the final position.

Worked example 3: two initial conditions

Question. A particle has acceleration a=6t−4 m s−2a=6t-4\,\mathrm{m\,s^{-2}}. At t=0t=0, its velocity is 3 m s−13\,\mathrm{m\,s^{-1}} and its displacement from OO is 2 m2\,\mathrm{m}. Find v(t)v(t), s(t)s(t) and its position at t=2t=2.

Plan. Integrate twice, using the relevant initial condition after each integration.

v=3t2−4t+C1.v=3t^2-4t+C_1.

Since v(0)=3v(0)=3, C1=3C_1=3. Now integrate this whole velocity:

s=t3−2t2+3t+C2.s=t^3-2t^2+3t+C_2.

Since s(0)=2s(0)=2, C2=2C_2=2. Therefore s(2)=8−8+6+2=8s(2)=8-8+6+2=8.

Answer. v=3t2−4t+3v=3t^2-4t+3, s=t3−2t2+3t+2s=t^3-2t^2+3t+2; at t=2t=2 the particle is 8 m8\,\mathrm{m} in the positive direction from OO.

Check. Differentiating ss gives the stated vv, and differentiating vv gives the supplied acceleration. Substituting t=0t=0 recovers both initial conditions. Its displacement during the first two seconds is 8−2=6 m8-2=6\,\mathrm{m}.

The next problem starts with velocity. A definite integral finds the change in position directly, but the initial position still matters.

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