The gradient on a distance-time graph gives the velocity, therefore if an equation involves distance and time then it can be differentiated to give velocity.
Similarly, the gradient on a velocity-time graph gives acceleration. Differentiating a velocity-time graph will give acceleration.
A particle is moving between two nodes. The displacement of the particle is given by `s = t^3 - 2t^2-12t`. What is the velocity of the particle after 5 seconds?
| Displacement | `s` | `= t^3 - 2t^2 -12t` |
| `frac(text(d)s)(text(d)t)` = velocity | `frac(text(d)s)(text(d)t)` | `= 3t^2 - 4t -12` |
| `v` | `= 3t^2 - 4t -12` | |
| At 5 seconds | `= 3(5)^2 - 4(5) -12` | |
| `= 43 ms^(-1)` |
Answer: `43 ms^(-1)`
The displacement of a particle is given by `s=t^3 - t^2 - t`. Determine when the velocity will be zero.
| Displacement | s | `= t^3 - t^2 -t` |
| velocity = | `frac(text(d)s)(text(d)t)` | `= 3t^2 - 2t -1` |
| When is velocity zero? | `0 ` | `= 3t^2 - 4t -1` |
| Use solution for quadratic | `x` | `= frac(-b ± sqrt(b^2-4ac))(2a)` |
| `` | `x` | `= frac(-(-2) ± sqrt((-2)^2-4(3)(-1)))(2(3))` |
| `` | `x` | `= 1 ` |
Ignore the negative time value, as it makes no sense in this context.
Answer: After 1 second