Answer :
Answer:
1050 m
Explanation:
Kinematics
Kinematics is a sub-field of physics that studies classic motion. Variable-based equations can be used to find the values of the basic features of motion.
- initial velocity v₀
- final velocity vₓ
- time t
- acceleration a
- displacement Δx
The equations are
[tex]\Delta x = v_0t+\dfrac{1}{2}at^2[/tex]
[tex]v_x=v_0+at[/tex]
[tex]v_x^2=v_0^2+2a\Delta x[/tex]
[tex]v=\dfrac{\Delta x}{t}[/tex]
[tex]a=\dfrac{v_x-v_0}{t}[/tex]
[tex]\dotfill[/tex]
Distance (d) vs. Displacement (Δx)
Distance and displacement may be used interchangeably in conversation , but it's crucial to know their differences when solving physics problems.
Distance measures how much an object has moved, regardless of its initial and final position.
The path an object takes starting from its initial point and stopping at its end point will be included in its calculation of distance.
Displacement measures only the space an object has veled relative to its initial and final points. Think of it has the measurement of how much "progress" an object has made.
This means that if an object starts and ends at the same spot, its displacement is 0 despite moving.
There can be occurences where distance is the same as displacement like if an object travels linearly.
[tex]\hrulefill[/tex]
Solving the Problem
Understanding the Problem
We're told the description that the car moves in
- starts from rest and accelerates for 15 seconds
- travels at a constant speed for 25 seconds more
- slows down to a stop in 5 seconds.
To find the total distance it travels or in this case, displacement, (the problem's description doesn't mention the car moving backward, so it never zero or negative displacement) we can break down the car's travelled distance into three parts:
- beginning
- middle
- end
[tex]\dotfill[/tex]
Beginning
We know from the first part of the problem
- v₀ = 0
- a = 2
- t = 15
and we need to find Δx.
We have to choose a kinematics equation that includes all of these variables so that we can rearrange (if necessary) and find the displacement.
We can utilize
[tex]\Delta x = v_0t+\dfrac{1}{2}at^2[/tex],
so we just plug and evaluate.
[tex]\Delta x = (0)(15)+\dfrac{1}{2}(2)(15)^2=\bold{225m}[/tex]
[tex]\dotfill[/tex]
Middle
In the second part of the problem,
- constant speed -> a = 0
- t = 25.
We're not giving any explicit information like in the last part.
Recalling that this constant speed was set by the car's final speed at the end of the "beginning" portion, we solve for the final speed after the car's initial acceleration.
Using the same information from the first part we can find the final speed of the car
[tex]v_x=v_0+at[/tex]
[tex]v_x=0+(2)(15)=30[/tex]
Now that we know the car's speed and its time, we can find its travelled distance using
[tex]v=\dfrac{\Delta x}{t}[/tex].
[tex]vt=\Delta x[/tex]
[tex](30)(25)=\bold{750m}[/tex]
[tex]\dotfill[/tex]
End
In the last part of the problem
- vₓ = 0
- t = 5
- v₀ = 30 (from the "Middle" part).
Looking at our kinematics equations, we need to find one more variable to find its displacement: acceleration.
[tex]a=\dfrac{0-30}{5}=-6[/tex]
So,
[tex]\Delta x = v_0t+\dfrac{1}{2}at^2[/tex]
[tex]\Delta x = (30)(5)+\dfrac{1}{2}(-6)(5)^2=\bold{75m}[/tex]
Putting it All Together
So, the total distance (displacement) traveled by the car is
[tex]\text{Total distance}=\text{beginning}+\text{middle}+\text{end}[/tex]
[tex]\text{Total distance}=225+750+75=\boxed{1050m}[/tex].
