Velocity Under Constant Acceleration

The time origin

In every scenario of this course, Time is zero at the start of the scenario. Time only increases during a scenario. An object can reverse its direction of motion, but time does not reverse.

Velocity over time

The Velocity of an object states its speed and its direction of motion at one instant. In this chapter, the object moves along one straight line. So the velocity is a signed number: its absolute value is the speed, and its sign is the direction. The Acceleration is constant during the scenario. On one axis, the vector equation (1.1.6) gives the velocity at time $\explain{motion.time}{t}$ as the Launch speed plus the acceleration multiplied by $\explain{motion.time}{t}$:

$$\explain{motion.velocity}{v(t)} = \explain{motion.launch-speed}{v_0} + \explain{motion.acceleration}{a} \explain{motion.time}{t}$$

(1.2.1)

(1.2.1) is exact, not an approximation. The acceleration is defined as the rate of change of the velocity. If the acceleration is constant, integrating this definition over time gives (1.2.1) without an error term [1]. The same integration for a vector is in § 1.1.4.1.

Example 1.2.1 Speeding up and braking at a constant acceleration

Entering a highway

A car enters a straight on-ramp at a Launch speed of $5\ \text{m/s}$ and has a constant Acceleration of $2\ \text{m/s}^2$. Substituting these values and a time of $4\ \text{s}$ into (1.2.1) gives its velocity:

$$\begin{aligned} \explain{motion.velocity}{v(4)} &= 5\ \text{m/s} + 2\ \text{m/s}^2 \times 4\ \text{s}\\ &= 5\ \text{m/s} + 8\ \text{m/s}\\ &= 13\ \text{m/s} \end{aligned}$$

Braking to a stop

A cyclist moves at a Launch speed of $8\ \text{m/s}$ and brakes with a constant Acceleration of $-2\ \text{m/s}^2$. Substituting these values and a time of $3\ \text{s}$ into (1.2.1) gives its velocity:

$$\begin{aligned} \explain{motion.velocity}{v(3)} &= 8\ \text{m/s} + (-2\ \text{m/s}^2) \times 3\ \text{s}\\ &= 8\ \text{m/s} - 6\ \text{m/s}\\ &= 2\ \text{m/s} \end{aligned}$$

Each second of braking takes $2\ \text{m/s}$ off the velocity:

Time	Velocity
$0\ \text{s}$	$8\ \text{m/s}$
$1\ \text{s}$	$6\ \text{m/s}$
$2\ \text{s}$	$4\ \text{m/s}$
$3\ \text{s}$	$2\ \text{m/s}$

Point at a symbol, in the text or in an equation, to open its explanation. Click the symbol to keep the explanation open while you read.

Table 1.2.1The two worked examples at a glance. Each scenario's launch speed, acceleration and time, and the velocity they give.

Scenario	Launch speed $\text{(m/s)}$	Acceleration $\text{(m/s}^2\text{)}$	Time $\text{(s)}$	Velocity $\text{(m/s)}$
Entering a highway	5	2	4	13
Braking to a stop	8	-2	3	2

Table 1.2.1 compares the inputs and the results of the two examples in Example 1.2.1. Knowledge check 1.2.1 tests the same calculation.

Knowledge check 1.2.1 Velocity under constant acceleration

1. DirectQuestion 1Not answered

References

1. Ling et al., University Physics Volume 1 (2016). ch. 3, “Instantaneous Velocity and Speed”. https://openstax.org/details/books/university-physics-volume-1 draft ↩