1) Acceleration: 0, distance: 195 km
2) Acceleration: [tex]2 m/s^2[/tex], distance: 25 m
3) Acceleration: [tex]2 m/s^2[/tex], distance: 108 m
4) Acceleration: [tex]2.5 m/s^2[/tex], distance: 45 m
5) Find the speed-time graphs in attachment
Explanation:
1)
The acceleration of an object is given by:
[tex]a=\frac{v-u}{t}[/tex]
where
u is the initial velocity
v is the final velocity
t is the time elapsed
In this problem, the car is travelling at constant speed, so
[tex]v=u[/tex]
and so the acceleration is zero:
[tex]a=0[/tex]
And therefore, we can calculate the distance covered using the equation:
[tex]d=vt[/tex]
where
v = 65 km/h is the speed
t = 3 h is the time
Substituting,
[tex]d=(65)(3)=195 m[/tex]
2)
We calculate the acceleration using the equation:
[tex]a=\frac{v-u}{t}[/tex]
where here we have:
u = 0 is the initial velocity
v = 10 m/s is the final velocity
t = 5 s is the time elapsed
Substituting,
[tex]a=\frac{10-0}{5}=2 m/s^2[/tex]
The distance can be calculated using the suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
u = 0
t = 5 s
[tex]a=2 m/s^2[/tex]
Substituting,
[tex]s=0+\frac{1}{2}(2)(5)^2=25 m[/tex]
3)
Again, we calculate the acceleration using:
[tex]a=\frac{v-u}{t}[/tex]
where this time we have:
u = 12 m/s is the initial velocity
v = 24 m/s is the final velocity
t = 6 s is the time elapsed
Substituting,
[tex]a=\frac{24-12}{6}=2 m/s^2[/tex]
The distance is calculated using the suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
u = 12 m/s
t = 6 s
[tex]a=2 m/s^2[/tex]
Substituting,
[tex]s=(12)(6)+\frac{1}{2}(2)(6)^2=108 m[/tex]
4)
Again, we calculate the acceleration using:
[tex]a=\frac{v-u}{t}[/tex]
where this time we have:
u = 54 km/h = 15 m/s is the initial velocity
v = 0 is the final velocity
t = 6 s is the time elapsed
Substituting,
[tex]a=\frac{0-15}{6}=-2.5 m/s^2[/tex]
The distance is calculated using the suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
u = 15 m/s
t = 6 s
[tex]a=-2.5 m/s^2[/tex]
Substituting,
[tex]s=(15)(6)+\frac{1}{2}(-2.5)(6)^2=45 m[/tex]
5)
The speed-time graph of each situation is shown in attachment. We notice that:
1) In case 1, the car travels at constant speed: this means that the speed-time graph is a horizontal line at v = 65 km/h
2) In case 2, there is a constant acceleration, with the speed changing from 0 to 10 m/s in 5 seconds: this is represented by a line having a slope of 2 (the slope of a speed-time graph corresponds to the acceleration)
3) Similarly, there is a constant acceleration, with the speed changing from 12 m/s to 24 m/s in 6 seconds: this is represented by a line having a slope of 2
4) This time there is a deceleration (negative acceleration), so the slope of the line here is negative.
Learn more about acceleration:
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