Position, Velocity, and Acceleration

From position to velocity and acceleration

Example 1

A particle's position is

x(t) = t^3 - 15t^2 + 63t - 40

feet for

0 \le t \le 8

seconds. (a) Find

v(t)

and

a(t)

. (b) Find the times when the particle is at rest and when it's traveling left. (c) Find

v(1)

and

a(1)

, and interpret each with units.

(a)

v(t) = 3t^2 - 30t + 63

, which factors as

3(t - 3)(t - 7)

, and

a(t) = 6t - 30

(b)

v(t) = 0

t = 3

and

t = 7

, so the particle is at rest at those two instants. When

3 < t < 7

, the factor

t - 3

is positive and

t - 7

is negative. So

v(t) < 0

, and the particle travels left. On

0 \le t < 3

and

7 < t \le 8

the two factors have the same sign, so

v(t) > 0

there: it travels left only on

3 < t < 7

(c)

v(1) = 36

: at

t = 1

second, the particle is traveling right at

36

feet per second.

a(1) = -24

: at

t = 1

second, its velocity is decreasing at

24

feet per second per second.

The position x(t). It rises while v > 0, turns at t = 3 and t = 7, and falls in between.

The velocity v(t) = 3(t − 3)(t − 7): positive, then negative for 3 < t < 7, then positive again.

Reading a velocity graph

Example 2

The graph shows the velocity

v(t)

of a particle traveling along a line for

0 \le t \le 5

. The velocity is

0

only at

t = 1

and

t = 4

, and the graph has a horizontal tangent only at

t = 2

and

t = 4

The velocity crosses zero at t = 1 and touches zero from below at t = 4.

(a) Find the times when the particle is at rest. (b) Find the intervals on which it travels left. (c) Determine at which of the times in (a) the particle changes direction, and justify your answer. (d) Find the intervals on which its acceleration is positive.

(a) At

t = 1

and

t = 4

, where

v(t) = 0

(b) The velocity is negative for

1 < t < 4

and

4 < t \le 5

, so the particle travels left on those intervals. At

t = 4

it stops for an instant and then keeps going left.

(c) Only at

t = 1

, where

v

changes from positive to negative. At

t = 4

the graph touches zero from below without crossing, so the direction doesn't change.

(d) The acceleration is the slope of the velocity graph. It's positive where the graph rises, on

2 < t < 4

Toward or away from each other

For two particles on the same line, let

d(t) = x_B(t) - x_A(t)

. If

d(t) > 0

, particle

B

is to the right of

A

, and the distance between them is

d(t)

. Its derivative is

d'(t) = v_B(t) - v_A(t)

. While

d > 0

, the particles are getting farther apart when

d' > 0

and closer together when

d' < 0

Example 3

Particle

A

has position

x_A(t) = t^2 - 6t + 2

and particle

B

has position

x_B(t) = 5 - t

, both in meters, for

t \ge 0

seconds. Determine whether the particles are heading toward each other or away from each other (a) at

t = 1

and (b) at

t = 4

The velocities are

v_A(t) = 2t - 6

and

v_B(t) = -1

(a) At

t = 1

x_A = -3

and

x_B = 4

, so

B

is to the right of

A

. Both travel left, with

v_A = -4

and

v_B = -1

, so

d'(1) = -1 - (-4) = 3 > 0

. Particle

A

goes left faster than

B

does, so the particles are heading away from each other.

At t = 1: A at x = −3 traveling left at 4 m/s, B at x = 4 traveling left at 1 m/s. The gap is growing.

(b) At

t = 4

x_A = -6

and

x_B = 1

, so

B

is still to the right. Now

v_A = 2

and

v_B = -1

, so

A

travels right and

B

travels left, toward each other, and

d'(4) = -1 - 2 = -3 < 0

Vertical motion

Example 4

A ball is thrown upward from the roof of a building. Its height above the ground is

h(t) = -16t^2 + 32t + 48

feet,

t

seconds after it's thrown, with upward as the positive direction. (a) Find the time when the ball is at rest, and its height then. (b) Find the time it hits the ground, and its velocity at that moment. (c) Find its acceleration, and say what its sign means.

(a)

v(t) = h'(t) = -32t + 32

, which is

0

t = 1

. The ball is at rest at the top of its path,

h(1) = 64

feet above the ground. Before that

v > 0

and it rises; after that

v < 0

and it falls.

(b)

h(t) = -16(t - 3)(t + 1) = 0

t = 3

for

t \ge 0

. Then

v(3) = -64

: the ball hits the ground traveling downward at

64

feet per second.

(c)

a(t) = -32

feet per second per second for every

t

: gravity decreases the upward velocity at a constant rate.