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

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