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

Question #155449

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 ?


1
Expert's answer
2021-01-15T05:13:50-0500

Explanations & Calculations


  • At t=0\small t=0 he has passed the signpost 5m\small 5m & heading east at 15ms1\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 S=x0+u0t+12at2\small S= x_0+u_0t+\frac{1}{2}at^2 which is meant for a motion under constant acceleration.
  • Note that the S\small S is the total displacement measured eastwards from the signpost. East+\small East\rightarrow+
  • Therefore,

S=5m+15ms1(t)+12(4ms2)(t2)S=5+15t+2t2(1)\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 2s\small 2s is

St=2=5+15(2s)+2(2s)2=43m\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 V=u+at\small V= u+at (Derivating the equation (1) also yeilds the same relationship).

V=15ms1+4ms2(t)V=15+4tVt=2=15+4×2s=23ms1\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 25ms1\small 25ms^{-1} could be found,

25ms1=15ms1+4tt=2.5s\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,

St=2.5s=5+15(2.5)+2(2.5)2=55maway from the signpost\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}


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