# Numeracy, Maths and Statistics

Velocity, acceleration and distance (instantaneous rest) additional Mathematics 2021
Velocity, acceleration and distance (instantaneous rest) additional Mathematics 2021

The SUVAT equations from equations of motion can only be used when an object is moving with constant acceleration. When the acceleration varies, this is when we must use calculus.

The diagram above shows the relationship between acceleration $a$, velocity $v$ and displacement $x$.

A particle $P$ is moving along the $x$-axis. The displacement $x\mathrm{m}$ from $O$ can we written as a function of $t$ $x = t^5 – 12t^2 +9.$ Find the formula for acceleration.

To find acceleration from displacement, we must differentiate twice. Differentiating once gives $v = \dfrac{\mathrm{d}x}{\mathrm{d}t} = \left(5t^4 – 24t\right)\mathrm{ms^{-1} }.$ This is the equation for the velocity of the particle. Differentiating again gives us the formula for the acceleration of the particle $a = \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = \left(20t^3 – 24\right)\mathrm{ms^{-2} }.$

A particle $P$ is moving along the $x$-axis. The acceleration $a\mathrm{ms^{-2} }$ from $O$ can be written as a function of $t$ $a = t + 19.$ Suppose we know that when $t=0,$ we have $x=0$ and when $t=4$ we have $x=10$. Find the displacement when $t=10$.

First we can find the velocity by integrating the acceleration equation $v = \int t + 19 \:\mathrm{d}t = \dfrac{1}{2}t^2 + 19t + c.$ Now that we have the velocity we can integrate again to find the displacement $x = \int \dfrac{1}{2}t^2 + 19t + c \:\mathrm{d}t = \dfrac{1}{6}t^3 + \dfrac{19}{2}t^2 + ct + k.$ From the question we know that the particle has not moved ($x=0$) when $t=0$. We can use this information to find the value of $k$ by substituting $x=0$ and $t=0$ into the displacement equation above \begin{align} x &= \dfrac{1}{6}t^3 + \dfrac{19}{2}t^2 + ct + k,\\ 0 &= 0 + k,\\ \ k&=0. \end{align} Now, we also know that when $t=4,$ we have $x=10$. Substituting these values and setting $k=0$ in the displacement equation enables us to find the value of $c$. \begin{align} x &= \dfrac{1}{6}t^3 + \dfrac{19}{2}t^2 + ct,\\ 10 &= \dfrac{4^3}{6} + \dfrac{19\times 4^2}{2} + 4c,\\ 10& – \dfrac{32}{3} – 152 = 4c,\\ \ c &=-\dfrac{229}{6}. \end{align} Substituting this value of $c$ into the displacement equation gives us an equation in only $x$ and $t$ $x = \dfrac{1}{6}t^3 + \dfrac{19}{2}t^2 -\dfrac{190}{3}.$ Substituting in $t=10$, gives \begin{align} x &= \dfrac{1}{6}t^3 + \dfrac{19}{2}t^2 -\dfrac{190}{3},\\ &= \dfrac{1}{6}\times 10^3 + \dfrac{19}{2}\times 10^2 -\dfrac{190}{3},\\ &= 1053.33 \mathrm{m} \text{ (to }2\text{ d.p.)} \end{align}

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