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

The time takes Kathy to overtake Stan is 6.6 sec.

Step-by-step 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)=(a)/(2)t^2`

Put the value into the formula

`S_(k)=(5.37)/(2)t^2`....(I)

Distance covers by Stan

`S_(s)=(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)

`(5.37)/(2)t^2=(4.12)/(2)(t+0.94)^2`

`t=(0.94*√(4.12))/(√(5.37)-√(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 = (1)/(2)at^2</p><p>[tex]d = (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 + (1)/(2)a_r * t^2`

`1.82 = -3.873* t + (1)/(2)*1.25 * t^2`

by solving above equation we have

`t = 6.64 s`

so it will overtake after 6.64 s