Question

A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the second point is 15 m/s. (a) Calculate the speed at the first point. (b) Calculate the acceleration. (c) At what prior distance from the first point was the car at rest?

Answer 1

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Step-by-step explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

`Sf^2=So^2+(2*a*x)`

`Sf=So+(a*t)`

We are going to put our data:

`(15m/s)^2=So^2+(2*a*60m)`

`15m/s=So+(a*6s)`

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

`√((15m/s)^2-(2*a*60m))=So`

`15m/s-(a*6s)=So`

`√((15m/s)^2-(2*a*60m))=15m/s-(a*6s)`

`[√((15m/s)^2-(2*a*60m))]^(2)=[15m/s-(a*6s)]^(2)`

`(15m/s)^2-(2*a*60m)}=(15m/s)^(2)-2*(a*6s)*(15m/s)+(a*6s)^(2)`

`-120m*a=-180m*a+36s^(2)*a^(2)`

`0=120m*a-180m*a+36s^(2)*a^(2)`

`0=-60m*a+36s^(2)*a^(2)`

`0=(-60m+36s^(2)*a)*a`

`0=a1`

`(60m)/(36s^(2)) = a2`

`1.67m/s^(2)=a2`

If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

`Sf=So+(a*t)`

`15m/s=So+1.67m/s^2*6s`

`15m/s-(1.67m/s^2*6s)=So`

`4.98m/s=So`

For part c:

We are going to use:

`Sf^2=So^2+(2*a*x)`

`(4.98m/s)^2=0^2+(2*(1.67m/s^2)*x)`

`(24.80m^2/s^2)/(3.34m/s^2)=x`

`7.42m=x`