Respuesta :
We have constant-acceleration equations to apply to the two cars separately.
(a) Let the times of travel for Kathy and Stan be,
[tex]t_{s}[/tex] = [tex]t_{k}[/tex] + 1.20 seconds,
Both start from rest,
[tex]v_{k} = v_{s} = 0[/tex], so the expressions for the distance travelled are,
⇒ [tex]x_{k} = \frac{1}{2} a_{k} t_{k} ^{2}[/tex] ,
⇒ [tex]x_{k} = \frac{1}{2} (5.10 m/s^{2} ) t_{k} ^{2}[/tex] , and
⇒ [tex]x_{s} = \frac{1}{2}(3.50 m/s^{2} )(t_{k}+ 1.20s ) ^ {2}[/tex]
When, Kathy overtakes Stan, the two distances will be equal, Setting [tex]x_{k} = x_{s}[/tex] , gives,
⇒ [tex]\frac{1}{2} (5.10 m/s^{2} ) t_{k} ^{2}[/tex] = [tex]\frac{1}{2}(3.50 m/s^{2} )(t_{k}+ 1.20s ) ^ {2}[/tex] ,
This we simplify and write in the standard form of a quadratic as,
⇒ [tex]1.66 t_{k}^{2} - 8.26 t_{k} - 4.96 = 0[/tex],
We solve using the quadratic formula,
⇒ t = - b ± [tex]\sqrt{b^{2}-4ac}[/tex] / 2a,
⇒ t = - (- 8.26) ± [tex]\frac{\sqrt{(- 8.26)^{2} - 4(1.66)(-4.96)}} {2(1.66)}[/tex] ,
⇒ t = 5.5 seconds, and - 0.54 (which is negligible, so we will not take it as a value).
Hence, (a) It takes t = 5.5 seconds to Kathy to overtake Stan.
Only the positive root makes sense physically, because the overtake point must be after the starting point in time.
(b) Use the equation from part (a) for distance of travel,
⇒ [tex]x_{k} = \frac{1}{2} a_{k} t_k^{2}[/tex] ,
⇒ [tex]x_{k} = \frac{1}{2} (5.1) (5.5)^{2}[/tex] ,
⇒[tex]x_{k} = 77.13 m[/tex].
Hence, (b) Kathy travels 77 m distance before she catches Stan.
(c) Remembering that [tex]v_{k} = v_{s} = 0[/tex] , the final velocities will be:
⇒ [tex]v_{k} = a_{k} t_{k} = (5.1 m/s^{2}) (5.5 s )[/tex] = 28.05 m/s,
∴ (c) Speed of Kathy's car at the instant she overtakes Stan is 28.05 m/s, and,
(d) ⇒ [tex]v_{s} = a_{s} t_{s} = (3.44 m/s^{2}) (6.7s )[/tex] = 23 m/s.
Speed of Stan's car at the instant he is overtaken by Kathy is 23 m/s.
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