CLASSICAL-MECHANICS Β· Interactive Practice | Unit 3 Β· Video 3

From Velocity to Acceleration: The Same Limit, Applied Twice

IKey Formulas

Formula Name Meaning
aave=Ξ”vΞ”t=v(t+Ξ”t)βˆ’v(t)Ξ”ta_{\text{ave}} = \dfrac{\Delta v}{\Delta t} = \dfrac{v(t+\Delta t) - v(t)}{\Delta t} Average acceleration Secant-line slope on the vv–tt graph
a(t)=lim⁑Δtβ†’0Ξ”vΞ”t=dvdta(t) = \displaystyle\lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} Instantaneous acceleration Tangent-line slope on the vv–tt graph
a=d2xdt2a = \dfrac{d^2 x}{dt^2} Second derivative Differentiate position twice

SI unit of acceleration: mβ‹…sβˆ’2\text{m} \cdot \text{s}^{-2} (meters per second squared).

IIInteractive Visualizations

Visualization 1 β€” Secant slope approaching the tangent

For v(t)=0.5 t2v(t) = 0.5\,t^2 at t=2t = 2, does the average acceleration settle on one number as Ξ”tβ†’0\Delta t \to 0?

πŸ’‘ This is the same limit that turns average velocity into instantaneous velocity β€” now applied one level higher, to velocity itself.

Visualization 2 — The derivative chain x→v→ax \to v \to a

Differentiate position twice: velocity is a line of slope bb, and acceleration is the constant bb.

Position   x = xβ‚€ + Β½ b tΒ²

Velocity   v = b t

Acceleration   a = b

πŸ’‘ A negative bb tilts the velocity line downward β€” the object slows or reverses β€” yet its acceleration stays the same constant bb.

Visualization 3 β€” Speed is height, acceleration is slope

Can an object move fast yet have zero acceleration β€” or move slowly yet accelerate?

IIIQuiz Questions

Question 1

A velocity changes from v(t)=8v(t) = 8 m/s to v(t+Ξ”t)=20v(t+\Delta t) = 20 m/s over a time interval of Ξ”t=4\Delta t = 4 s. What is the average acceleration over this interval?

βœ… Correct! Ξ”v = 12 m/s over Ξ”t = 4 s gives 3 m/sΒ².

❌ Not quite. Use a_ave = Ξ”v/Ξ”t = (20 βˆ’ 8)/4. Don't forget to divide by Ξ”t, and use the change in velocity, not a single value.

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Solution:

Average acceleration is the change in velocity divided by the elapsed time:

aave=Ξ”vΞ”t=v(t+Ξ”t)βˆ’v(t)Ξ”t=20βˆ’84=124=3Β m/s2.a_{\text{ave}} = \frac{\Delta v}{\Delta t} = \frac{v(t+\Delta t) - v(t)}{\Delta t} = \frac{20 - 8}{4} = \frac{12}{4} = 3 \ \text{m/s}^2.

The answer is 3 m/sΒ².

Question 2

A jet is cruising in a straight line at a steady 900 km/h. While its speed stays constant, what is its acceleration?

βœ… Correct! Constant velocity means Ξ”v = 0, so a = dv/dt = 0 β€” no matter how fast the object is going.

❌ Not quite. This is the classic 'fast = accelerating' trap. Acceleration is the rate of change of velocity, not the velocity itself. Constant speed in a straight line means a = 0.

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Solution:

Acceleration measures how velocity changes, not how large velocity is:

a=dvdt.a = \frac{dv}{dt}.

If the velocity is constant (steady speed, straight line), then Ξ”v=0\Delta v = 0 over any interval, so a=0a = 0. A fast object can have zero acceleration; a slow object can have large acceleration. Speed and acceleration are independent.

The answer is zero.

Question 3

For the worked example, the velocity is v(t)=b tv(t) = b\,t where bb is a constant. Using the limit definition a(t)=lim⁑Δtβ†’0v(t+Ξ”t)βˆ’v(t)Ξ”ta(t) = \lim_{\Delta t \to 0}\dfrac{v(t+\Delta t) - v(t)}{\Delta t}, what is the instantaneous acceleration a(t)a(t)?

βœ… Correct! The Ξ”t factors cancel and b is constant, so the limit is just b β€” constant acceleration.

❌ Not quite. After cancelling the bΒ·t terms, the quotient is bΒ·Ξ”t/Ξ”t = b, and the Ξ”t's cancel. The result does not depend on t.

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Solution:

Apply the limit definition with v(t)=b tv(t) = b\,t:

a(t)=lim⁑Δtβ†’0v(t+Ξ”t)βˆ’v(t)Ξ”t.a(t) = \lim_{\Delta t \to 0} \frac{v(t+\Delta t) - v(t)}{\Delta t}.

Evaluate vv at t+Ξ”tt + \Delta t:

v(t+Ξ”t)=b (t+Ξ”t)=b t+b Δt.v(t+\Delta t) = b\,(t + \Delta t) = b\,t + b\,\Delta t.

Form the difference quotient (the b tb\,t terms cancel):

(b t+b Δt)βˆ’b tΞ”t=b ΔtΞ”t=b.\frac{(b\,t + b\,\Delta t) - b\,t}{\Delta t} = \frac{b\,\Delta t}{\Delta t} = b.

Since bb does not depend on Ξ”t\Delta t, the limit is simply bb:

a(t)=b(constantΒ acceleration).a(t) = b \quad (\text{constant acceleration}).

The answer is a(t) = b.

Question 4

True or False: Acceleration is the second derivative of position with respect to time, so a=d2xdt2a = \dfrac{d^2 x}{dt^2}.

βœ… Correct! a = dv/dt and v = dx/dt, so a = dΒ²x/dtΒ² β€” the same limit applied twice.

❌ Not quite. Velocity is the first derivative of position and acceleration is the derivative of velocity, so acceleration is the second derivative of position. The statement is True.

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Solution:

Velocity is the first derivative of position:

v=dxdt.v = \frac{dx}{dt}.

Acceleration is the derivative of velocity:

a=dvdt.a = \frac{dv}{dt}.

Substituting the first relation into the second, acceleration is the derivative of a derivative of position β€” the second derivative:

a=ddt ⁣(dxdt)=d2xdt2.a = \frac{d}{dt}\!\left(\frac{dx}{dt}\right) = \frac{d^2 x}{dt^2}.

So the statement is True. Differentiate position once for velocity, twice for acceleration.

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