SUVAT

Key Stage 4

Meaning

A suvat is an equation of motion for an object.

About suvat Equations

suvat equations can be used to find one of these variables:

• Displacement (s) - How far an object is from its starting position.
• Initial Velocity (u) - The velocity of an object at the start.
• Final Velocity (v) - The velocity of an object at the end.
• Acceleration (a) - The rate of change of velocity.
• Time (t) - The time taken between to points on a journey.

Equations

\(v=\frac{s}{t}\)

\(a=\frac{v-u}{t}\)

\(v^2=u^2 + 2as\)

Example Calculations

Using v, s and t

A wave travels 1000m in a time of 12.5s. Calculate the magnitude of the velocity of the wave correct to two significant figures.	The Moon is approximately 390,000km away from the Earth. A Radio Wave travels at 300,000,000m/s from The Moon to the Earth. Calculate the time taken by the Radio Wave to reach Earth correct to two significant figures.	An alpha particle travels for 4.0ns at a velocity of 15,000,000m/s before colliding with an air molecule. Calculate the displacement of the alpha particle from the start to the end of its journey correct to two significant figures.
1. State the known quantities s = 1000m t = 12.5s	1. State the known quantities s = 390,000km = 390,000,000m v = 300,000,000m/s	1. State the known quantities v = 15,000,000m/s t = 4.0ns = 4.0 x 10-9m
2. Substitute the numbers and evaluate. \(v = \frac{s}{t}\) \(v = \frac{1000}{12.5}\) \(v = 80m/s\)	2. Substitute the numbers and evaluate. \(v = \frac{s}{t}\) \(300000000 = \frac{390000000}{t}\)	2. Substitute the numbers and evaluate. \(v = \frac{s}{t}\) \(15000000 = \frac{s}{4.0 \times 10^{-9}}\)
3. Rearrange the equation and solve. Already solved.	3. Rearrange the equation and solve. \(300000000 = \frac{390000000}{t}\) \(300000000t = 390000000\) \(t = \frac{390000000}{300000000}\) \(t = 1.3s\)	3. Rearrange the equation and solve. \(15000000 = \frac{s}{4.0 \times 10^{-9}}\) \(s = 15000000 \times 4.0 \times 10^{-9}\) \(s = 0.060m\)

Using a, v, u and t

A cyclist travels at an initial velocity of 5m/s. They then take 20 seconds to increase velocity to 15m/s. Calculate the acceleration of the cyclist correct to two significant figures.	A driverless car travelling at 35m/s notices a pedestrian walk into the road and begins automatic breaking. It comes to a complete stop. The acceleration is -14m/s. Calculate the time taken to come to a stop correct to two significant figures.
1. State the known quantities u = 5m/s v = 15m/s t = 20s	1. State the known quantities a = -14m/s/s u = 35m/s v = 0m/s
2. Substitute the numbers and evaluate. \(a=\frac{v-u}{t}\) \(a=\frac{15-5}{20}\) \(a=\frac{10}{20}\)	2. Substitute the numbers and evaluate. \(a=\frac{v-u}{t}\) \(-14=\frac{0-35}{t}\) \(-14=\frac{-35}{t}\)
3. Rearrange the equation and solve. \(a=0.50m/s/s\)	3. Rearrange the equation and solve. \(-14t=-35\) \(t=\frac{-35}{-14}\) \(t = 2.5s\)

A person runs onto a thick sheet of ice. They begin to slide along. They decelerate at a rate of 0.2m/s/s for 30 seconds until they come to rest. Calculate the initial velocity of person correct to two significant figures.	A safety car is out on the track with F1 cars limited to 15m/s until it pulls over. A car then accelerates at 9m/s/s for 8 seconds. Calculate the final velocity of the car correct to two significant figures.
1. State the known quantities a = -0.2m/s/s v = 0m/s t = 30s	1. State the known quantities a = 9m/s/s u = 15m/s t = 8s
2. Substitute the numbers and evaluate. \(a=\frac{v-u}{t}\) \(-0.2=\frac{0-u}{30}\) \(-0.2=\frac{-u}{30}\)	2. Substitute the numbers and evaluate. \(a=\frac{v-u}{t}\) \(9=\frac{v-15}{8}\)
3. Rearrange the equation and solve. \(-0.2=\frac{-u}{30}\) \(-u= (-0.2) \times 30\) \(u= 6.0m/s\)	3. Rearrange the equation and solve. \(9 \times 8 =v-15\) \(9 \times 8 + 15 = v\) \(v =87m/s\)

Using v, u a and s

A fighter jet accelerates at 60m/s/s from rest along a runway which is 150m in length. Calculate the velocity reached by the fighter jet, correct to two significant figures.	A train applies the breaks over 25m of track slowing at a rate of 0.5m/s/s to a velocity of 10m/s. Calculate the initial velocity of the train prior to breaking, correct to two significant figures.
1. State the known quantities u = 0m/s a = 60m/s/s s = 150m	1. State the known quantities v = 10m/s a = -0.5m/s/s s = 25m
2. Substitute the numbers and evaluate. \(v^2=u^2 + 2as\) \(v^2=0^2 + 2 \times 60 \times 150\) \(v^2=18000\)	2. Substitute the numbers and evaluate. \(v^2=u^2 + 2as\) \(10^2=u^2 + 2 \times (-0.5) \times 25\) \(100=u^2 - 25\)
3. Rearrange the equation and solve. \(v=\sqrt{18000}\) \(v=134.16m/s\) \(v \approx 130m/s \)	3. Rearrange the equation and solve. \(u^2 = 100 + 25\) \(u^2 = 125\) \(u=\sqrt{125}\) \(u=11.18m/s\) \(u \approx 11m/s\)

A car travelling at 30m/s has a stopping distance of 55m. Calculate the deceleration of the car correct to two significant figures.	A formula 1 car is travelling at 50m/s on a straight but must slow to 22m/s for an approaching corner. The breaks can decelerate the car at 30m/s/s. Calculate how far before the corner the car must begin braking, correct to two significant figures.
1. State the known quantities v = 0m/s u = 30m/s s = 55m	1. State the known quantities v = 22m/s u = 50m/s a = -30m/s/s
2. Substitute the numbers and evaluate. \(v^2=u^2 + 2as\) \(0^2=30^2 + 2 \times a \times 55\) \(0=900+110a\)	2. Substitute the numbers and evaluate. \(v^2=u^2 + 2as\) \(22^2=50^2 + 2 \times (-30) \times s\) \(484=2500 - 60s\)
3. Rearrange the equation and solve. \(110a = -900\) \(a =\frac{-900}{110}\) \(a =\frac{-900}{110}\) \(a=-8.1818m/s/s\) \(a \approx -8.2m/s/s \) Deceleration = + 8.2m/s/s	3. Rearrange the equation and solve. \(484=2500 - 60s\) \(60s = 2500-484\) \(s =\frac{2016}{60}\) \(s=33.6m\) \(s \approx 34m\)