Question

A young woman named kathy kool buys a sports car that can accelerate at the rate of 5.37 m/s 2 . she decides to test the car by drag racing with another speedster, stan speedy. both start from rest, but experienced stan leaves the starting line 0.94 s before kathy. stan moves with a constant acceleration of 4.12 m/s 2 and kathy maintains an acceleration of 5.37 m/s 2 . find the time it takes kathy to overtake stan. answer in units of s.

Answer 1

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

$S_{k}=\dfrac{a}{2}t^2$

Put the value into the formula

$S_{k}=\dfrac{5.37}{2}t^2$....(I)

Distance covers by Stan

$S_{s}=\dfrac{4.12}{2}(t+0.94)^2$....(II)

When Kathy to overtake Stan then the distance of Kathy and Stan should be equal

From equation (I) and (II)

$\dfrac{5.37}{2}t^2=\dfrac{4.12}{2}(t+0.94)^2$

$t=\dfrac{0.94\times\sqrt{4.12}}{\sqrt{5.37}-\sqrt{4.12}}$

$t=6.6\ sec$

Hence, The time takes Kathy to overtake Stan is 6.6 sec.  

Answer 2

As the stan moves 0.94 s before with an acceleration 4.12 m/s^2

so the distance moved by it

$d = \frac{1}{2}at^2[tex]d = \frac{1}{2}*4.12*0.94^2$

$d = 1.82 m$

speed gained by the car is given as

$v = v_i + at$

$v = 4.12* 0.94 = 3.873 m/s$

now the relative speed of them is given as

$v_r = 3.873 - 0 = 3.873 m/s$

relative acceleration is given as

$a_r = 4.12 - 5.37 = -1.25$

now the distance between them is to be covered

$d = v_r * t + \frac{1}{2}a_r * t^2$

$1.82 = -3.873* t + \frac{1}{2}*1.25 * t^2$

by solving above equation we have

$t = 6.64 s$

so it will overtake after 6.64 s