Respuesta :
As the stan moves 0.94 s before with an acceleration 4.12 m/s^2
so the distance moved by it
[tex]d = \frac{1}{2}at^2
[tex]d = \frac{1}{2}*4.12*0.94^2[/tex]
[tex]d = 1.82 m[/tex]
speed gained by the car is given as
[tex]v = v_i + at[/tex]
[tex]v = 4.12* 0.94 = 3.873 m/s[/tex]
now the relative speed of them is given as
[tex]v_r = 3.873 - 0 = 3.873 m/s[/tex]
relative acceleration is given as
[tex]a_r = 4.12 - 5.37 = -1.25[/tex]
now the distance between them is to be covered
[tex]d = v_r * t + \frac{1}{2}a_r * t^2[/tex]
[tex]1.82 = -3.873* t + \frac{1}{2}*1.25 * t^2[/tex]
by solving above equation we have
[tex]t = 6.64 s[/tex]
so it will overtake after 6.64 s
Answer:
The time takes Kathy to overtake Stan is 6.6 sec.
Explanation:
Given that,
Acceleration of car = 5.37 m/s²
Time = 0.9 sec
Acceleration of Stan's car = 4.12 m/s²
We need to calculate the distance covers by Kathy
Using equation of motion
[tex]S_{k}=\dfrac{a}{2}t^2[/tex]
Put the value into the formula
[tex]S_{k}=\dfrac{5.37}{2}t^2[/tex]....(I)
Distance covers by Stan
[tex]S_{s}=\dfrac{4.12}{2}(t+0.94)^2[/tex]....(II)
When Kathy to overtake Stan then the distance of Kathy and Stan should be equal
From equation (I) and (II)
[tex]\dfrac{5.37}{2}t^2=\dfrac{4.12}{2}(t+0.94)^2[/tex]
[tex]t=\dfrac{0.94\times\sqrt{4.12}}{\sqrt{5.37}-\sqrt{4.12}}[/tex]
[tex]t=6.6\ sec[/tex]
Hence, The time takes Kathy to overtake Stan is 6.6 sec.
