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padilas [110]
3 years ago
5

How do you solve this problem?

Physics
1 answer:
Katarina [22]3 years ago
7 0

The particle has acceleration vector

\vec a=\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\jmath

We're told that it starts off at the origin, so that its position vector at t=0 is

\vec r_0=\vec0

and that it has an initial velocity of 12 m/s in the positive x direction, or equivalently its initial velocity vector is

\vec v_0=\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\,\vec\imath

To find the velocity vector for the particle at time t, we integrate the acceleration vector:

\vec v=\vec v_0+\displaystyle\int_0^t\vec a\,\mathrm d\tau

\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath

\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath

Then we integrate this to find the position vector at time t:

\vec r=\vec r_0+\displaystyle\int_0^t\vec v\,\mathrm d\tau

\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath

\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath

Solve for the time when the y coordinate is 18 m:

18\,\mathrm m=\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\implies t=3.0\,\mathrm s

At this point, the x coordinate is

\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)(3.0\,\mathrm s)+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3.0\,\mathrm s)^2=27\,\mathrm m

so the answer is C.

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\blue{\bold{\underline{\underline{Answer:}}}}

\green{\tt{\implies{Time=10\:sec\:and\:30\:sec}}} \\\\

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}} \\ \\

\green{\underline{\bold{Given :}}}  \\ \\   \:\:\:\: \bullet\:\:\tt\red{ Velocity \: of \: boy = 50 \: m/s} \\  \\   \:\:\:\: \bullet\:\: \tt\orange{Velocity \: of \: girl = 30 \: m/s }\\  \\  \:\:\:\: \bullet\:\:\tt\green{ Acceleration \: of \: boy =  {1 \: m/s}^{2}} \\  \\  \:\:\:\:\bullet\:\: \tt\blue{Acceleration \: of \:girl=  {2\: m/s}^{2} }\\  \\ \:\:\:\: \bullet \:\:\tt\purple{Sepration \: between \: them = 150 \: m }\\ \\  \\ \red{\underline{\bold{To \: Find :}}} \\ \\  \:\:\:\: \bullet\:\: \tt\blue{Time \: taken \: to \: catch \: the \: girl  }\\

<u>According to given question</u> :

\\ \green{\star} \:  \text{Using \: relative \: motion \: method} \\  \\  \green{ \circ} \:  \tt Net \: velocity = 50 - 30 = 20 \: m/s \\  \\ \green{ \circ} \:  \tt Net \: acceleration = 1 - 2 = - 1\: m /{s}^{2}  \\\\  \\  \star\:\bold\red{\underline{\:As \: we \: know \: that\:}} \\\\  \tt\purple{:  \implies s = ut +  \frac{1}{2}  {at}^{2}}  \\  \\ \tt\green{:  \implies 150 = 20 \times t +  \frac{1}{2}  \times  -1 \times  {t}^{2}}  \\  \\ \tt\purple{:  \implies 300 = 40t -  {t}^{2}}  \\  \\ \tt\green{:  \implies  {t}^{2}  - 40t  + 300 = 0} \\  \\ \tt\purple{:  \implies t =  \frac{  - ( - 40) \pm\sqrt{ { (- 40)}^{2}  - 4 \times 1 \times 300} }{2 \times 1}  }\\  \\ \tt\green{:  \implies t =  \frac{40 \pm \sqrt{1600 - 1200} }{2}  }\\  \\ \tt\purple{:  \implies t =  \frac{40 \pm 20}{2}  }\\  \\  \green{\tt:  \implies t = 10 \: sec \: and \: 30 \: sec}

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