Differentiation rules practice question 16
A particle moves along a line, and its position at time seconds is metres. Which describes its motion at ?
The direction of motion is the sign of the velocity , not the sign of the position.
Speed is . It increases when and have the same sign, and decreases when their signs differ.
Here and . Evaluate both at and compare their signs.
Theory UBC Math 100 §3.4CLP-1 §2.14Key results
Worked solution
Strategy
Compute the velocity and the acceleration at t = 3. The sign of v gives the direction; comparing the signs of v and a says whether the speed is growing or shrinking.
Working
Negative velocity: the particle is moving in the negative direction.
Positive acceleration: the velocity is increasing.
The velocity rises from −9 toward 0, so the speed falls: the particle is slowing down.
Answer
The answer is moving in the negative direction and slowing down.
Check
From t = 2.9 to 3 the position drops 0.929 m, and from 3 to 3.1 only 0.869 m: it moves in the negative direction, and more slowly.
Where this goes wrong
Reading a positive acceleration as speeding up. Acceleration is the rate of change of velocity, not of speed: with v = −9 and a = 6 the velocity climbs toward 0 and the particle slows.
The idea behind it
The differentiation rules lesson, a figure to move, in a few steps
Try a similar question
A particle moves along the -axis with position at time . Find its acceleration at .Saved in this browser only.