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

Answer:

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

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.

$\sqrt{(15m/s)^2-(2*a*60m)}=So$

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

$\sqrt{(15m/s)^2-(2*a*60m)}=15m/s-(a*6s)$

$[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}$

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

$-120m*a=-180m*a+36s^{2}*a^{2}$

$0=120m*a-180m*a+36s^{2}*a^{2}$

$0=-60m*a+36s^{2}*a^{2}$

$0=(-60m+36s^{2}*a)*a$

$0=a1$

$\frac{60m}{36s^{2}} = a2$

$1.67m/s^{2}=a2$

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:

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

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

$\frac{24.80m^2/s^2}{3.34m/s^2}=x$

$7.42m=x$