← Back to learning objective

Rewrite piecewise and find velocity for t < 3

A particle moves along a line with position function s(t)=∣t−3∣+2ts(t) = |t-3| + 2t, where ss is measured in meters and tt in seconds, for t≥0t \geq 0.

For t<3t < 3, the expression ∣t−3∣|t-3| simplifies to 3−t3-t, so s(t)=3−t+2t=t+3s(t) = 3 - t + 2t = t + 3. Find the velocity v(t)=s′(t)v(t) = s'(t) for t<3t < 3.

Hint

Since t−3<0t - 3 < 0 when t<3t<3, we have ∣t−3∣=−(t−3)=3−t|t-3| = -(t-3) = 3-t. Simplify s(t)s(t) into a single linear expression, then differentiate.

Step-by-step solution

For t<3t<3, t−3t - 3 is negative, so ∣t−3∣=−(t−3)=3−t|t-3| = -(t-3) = 3-t. Substituting gives s(t)=(3−t)+2t=t+3s(t) = (3-t) + 2t = t+3.

Differentiating this linear function gives s′(t)=1s'(t) = 1. Thus the velocity for t<3t<3 is constant: v(t)=1v(t) = 1 m/s.

Find velocity for t > 3

For t>3t > 3, the expression ∣t−3∣|t-3| simplifies to t−3t-3, so s(t)=(t−3)+2t=3t−3s(t) = (t-3) + 2t = 3t - 3. Find the velocity v(t)=s′(t)v(t) = s'(t) for t>3t > 3.

Hint

Since t−3>0t-3 > 0 when t>3t>3, ∣t−3∣=t−3|t-3| = t-3. Combine like terms in s(t)s(t) before differentiating.

Step-by-step solution

For t>3t>3, t−3t-3 is positive, so ∣t−3∣=t−3|t-3| = t-3. Substituting gives s(t)=(t−3)+2t=3t−3s(t) = (t-3) + 2t = 3t - 3.

Differentiating this linear function gives s′(t)=3s'(t) = 3. Thus the velocity for t>3t>3 is constant: v(t)=3v(t) = 3 m/s.

Differentiability at t = 3

Based on the velocities found for t<3t<3 and t>3t>3, is s(t)s(t) differentiable at t=3t = 3? Justify your answer by comparing the left-hand and right-hand derivatives at t=3t=3.

Hint

A function is differentiable at a point only if the left-hand and right-hand derivatives exist and are equal there. Compare the two constant velocities found in the previous parts.

Step-by-step solution

From the previous parts, the left-hand derivative as t→3−t \to 3^- is 11, while the right-hand derivative as t→3+t \to 3^+ is 33.

Since these one-sided derivatives are not equal (1≠31 \neq 3), the derivative s′(3)s'(3) does not exist. Therefore s(t)s(t) is not differentiable at t=3t=3; the graph has a corner there where the velocity jumps abruptly from 1 to 3 m/s.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap