Having problems with Physics.

Question

A car moving with a speed of 108 km/h is brought to rest in a distance of 60 m. The car's acceleration is -7.5 m/s/s. How much time did the car take to stop?

Answer 1

Rachel,

Both solutions show you the correct answer; but they show you to plug in the values before you do the algebra to solve for time.

(besides there is a much easier way - see below)

 

You should Never plug in the values first.

You should Always solve the equation for the unknown first.

 

I can show you studies that show students make more mistakes when they plug in values first, and I know of no physics or science teacher that teaches problem solving that way.

 

When you plug in values first students do not see what variables cancel out, they will take out a calculator before solving for the unknown, then they will try to do the algebra in their head while trying to punch things into the calculator (and make mistakes), and they will not be able to do a quick unit analysis to see if things are correct. They make more mistakes.

 

If you notice both the solutions put numbers into the equation without units.  That is wrong.  Always put units with all values.  Otherwise how do you know the units of the answer will come out right.

 

If you solve the problem for time (t) 1st you get this:

Δt = -v_o ± √(v_o² + 2aΔx)

                 a

 

ALWAYS solve for the unknown FIRST

Then put in Values (numbers With units)

Then you can see if the units are correct with unit analysis

If the units are right you have everything correct.

NOW take out your calculator.

 

You will make less mistakes that way.

I guarantee it.

 

--------EASIER WAY---------------

Also there is a much easier way to solve this.

 

a = Δv/Δt

Δt - Δv/a

Δv = v_f - v_o = (0 - 108 km/hr)

Δt= (0 - 108 km/hr)/ -7.5 m/s^2

 

Answer 2

Rachel,

 

The standard formula for distance traveled under constant acceleration is:

 

X(t) = X_o + V_o t + (1/2) a t²  , where

     

X(t) = distance traveled in time t (given here as 60 m).

X_o   = starting position at time t=0 (take as zero here, the origin of our coordinate system for X).

V_o   = starting velocity at time t=0 (given here as 108 km/h).

a     = constant acceleration (or deceleration when negative) over time t (given here as - 7.5 m/sec²)

t*    = time required to bring car to a complete stop (to be computed here).    

 

So plug in the values and solve for t = t*.

 

60 = 0 + 30 t  + (1/2) (-7.50 t² ).   Remember to change Vo to m/sec as it's given in Km/hr

                                                            108000m/hr (1/3600 sec/hr) = 30 m/sec.

3.75 t² - 30 t + 60 = 0,   you can use the quadratic formula to solve for t, where A=3.75, B=-30 & C=60

 

t  = {[-B ± √(B² - 4AC)]/ 2A}  = { [30 ± √(900 - 900)]/ 7.5} = 30 / 7.5  = 4.0

 

Thus, t* = 4.0 seconds to stop that car.

 

Answer 3

Hello Rachel,

 

We can apply the following, assuming these variables from the word problem:

 

Initial Position of the Car = 0m = x₀

Initial Velocity of the Car = 108km/h = v₀

Constant Acceleration of the Car = -7.5m/s^2 = a

Final Position of the Car = 60m = x_f

 

We can then solve the problem using the following equation:

 

x_f = x₀ + v₀t + 0.5at²

 

60m = 0m + (108 km/h)*t + 0.5*(-7.5 m/s²)*t²

 

converting all of the units to base units:

 

60m = 0m + (108 km/h)*(1000m/1km)*(1h/3600s)*t + 0.5*(-7.5 m/s²)*t²

 

simplifying and rearranging the equation in the form:

 

ax2+bx+c = 0

 

-3.75t² + 30t - 60 = 0

 

Solving for t, using the quadratic formula:

 

[-b ± sqrt(b²-4ac)]/2a = x 

 

[-30 ± sqrt((30)² - 4*(-3.75)*(-60))]/(2*-3.75) = t

 

t = 4 ± sqrt(900-900)/(-7.5) = 4 ± 0 = 4

 

Therefore, it takes 4 seconds for the car to stop.