SUVAT
Contents
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.
\(484=2500 - 60s\) \(60s = 2500-484\) \(s =\frac{2016}{60}\) \(s=33.6m\) \(s \approx 34m\) |