Answer to Question #155449 in Mechanics | Relativity for Abdul Raffih

Question

a motorcyclist heading east through small city Acelerates after he The sign post marking the city limits. this acceleration is a Constant 4.0m/s² at time t=0 he is 5.0 m east of the signpost tmoving east at 15 m / s.

Find the position and velocity at time t = 2.0s? when is the motorcyclist when his velocity id 25 / s ?

Answer 1

Explanations & Calculations

• At $\small t=0$ he has passed the signpost $\small 5m$ & heading east at $\small 15ms^{-1}$.
• It is quite straight forward to calculate the displacement of him at a given time on his travel (under the given acceleration) here onwards, using the equation $\small S= x_0+u_0t+\frac{1}{2}at^2$ which is meant for a motion under constant acceleration.
• Note that the $\small S$ is the total displacement measured eastwards from the signpost. $\small East\rightarrow+$
• Therefore,

$\qquad\qquad \begin{aligned} \small S &= \small 5m+15ms^{-1}( t)+\frac{1}{2}(4ms^{-2})(t^2)\\ \small S &= \small 5+15t+2t^2\cdots(1) \end{aligned}$

• Then the position at $\small 2s$ is

$\qquad\qquad \begin{aligned} \small S_{t=2} &= \small 5+15(2s)+2(2s)^2\\ &= \small \bold{43m} \end{aligned}$

• Velocity at a given time can be calculated using the equation $\small V= u+at$ (Derivating the equation (1) also yeilds the same relationship).

$\qquad\qquad \begin{aligned} \small V&= 15ms^{-1}+4ms^{-2}(t)\\ \small V&= 15+4t\\ \small V_{t=2}&= \small 15+4\times 2s\\ &= \small \bold{23ms^{-1}} \end{aligned}$

• Using the same equation, the time when the velocity becomes $\small 25ms^{-1}$ could be found,

$\qquad\qquad \begin{aligned} \small 25ms^{-1}&=\small 15ms^{-1}+4t\\ \small t&= \small \bold{2.5s} \end{aligned}$

• Plugging this in equation (1), the position could be calculated,

$\qquad\qquad \begin{aligned} \small S_{t=2.5s}&= \small 5+15(2.5)+2(2.5)^2\\ &= \small \bold{55m\quad\text{away from the signpost}} \end{aligned}$