# 2.2.1 Equations of Motion

Uniform acceleration motion refers to rectilinear motion with constant acceleration. If we are provided with the initial velocity u and the (constant) acceleration a, we can calculate the subsequent velocity v and displacement s at any time t using the following three equations.

v−t equation: $v=u+at$

s−t equation: $s=ut+\frac{1}{2}a{{t}^{2}}$

v−s equation: ${{v}^{2}}={{u}^{2}}+2as$

Collectively, they are called the equations of motion, a.k.a. the suvat equations. Individually, I call them the v-t, s-t and v-s equations. Notice that the symbols in red are constants, while those in blue are not.

Example 1

From rest, a car accelerates at a constant rate of 2.0 m s-2. Calculate

1. its velocity 3.0 seconds later,
2. its position 3.0 seconds later,
3. its velocity 3.0 m from the starting position.

Solution

1. Since we are working for v at particular t, we use the v-t equation. \begin{aligned}(v&=u+at)\\v&=0+(2.0)(3.0)\\&=6.0\text{ m }{{\text{s}}^{{-1}}}\end{aligned}

• Since we are working for s at particular t, we use the s-t equation. \begin{aligned}(s&=ut+\frac{1}{2}a{{t}^{2}})\\s&=0+\frac{1}{2}(2.0){{(3.0)}^{2}}\\&=9.0\text{ m}\end{aligned}

• Since we are working for v at particular s, we use the v-s equation. \begin{aligned}({{v}^{2}}&={{u}^{2}}+2as)\\{{v}^{2}}&=0+2(2.0)(3.0)\\v&=3.46\text{ m }{{\text{s}}^{{-1}}}\end{aligned}

Example 2

The velocity of a car at $t=0\text{ s}$ is 8.0 m s-1 rightward. It has a constant acceleration of 3.0 m s-2 in the leftward direction. Calculate

1. its velocity 2.0 seconds later,
2. its position 6.0 seconds later,
3. the furthest to the right reached by the car.

Solution

1. Since we are working for v at particular t, we use the v-t equation. \displaystyle \begin{aligned}(v&=u+at)\\v&=8.0+(-3.0)(2.0)\\&=2.0\text{ m }{{\text{s}}^{{-1}}}\end{aligned}

Notice that we are taking the sign convention of rightward being +ve. That’s why we write u as positive 8.0 but a as negative 3.0.

• Since we are working for s at particular t, we use the s-t equation. \begin{aligned}(s&=ut+\frac{1}{2}a{{t}^{2}})\\s&=(8.0)(6.0)+\frac{1}{2}(-3.0){{(6.0)}^{2}}\\&=-6.0\text{ m}\end{aligned}

Since our sign convention is rightward being +ve, the negative sign implies that the car’s position is 6.0 m to the left of the initial position.

• Since we are working for s at particular v, we use the v-s equation. \begin{aligned}({{v}^{2}}&={{u}^{2}}+2as)\\{{0}^{2}}&={{(8.0)}^{2}}+2(-3.0)s\\s&=10.7\text{ m}\end{aligned}

Notice that the car travels rightward as long as the velocity is positive. It turns back only after its velocity turns negative. That’s why to solve for the furthest distance to the right is to solve for s at $v=0$.