How Does an Object's Acceleration Affect its Initial Velocity?

[negation]

Homework Statement

An object undergoes acceleration 2.3i + 3.6j for 10s. At the end of this time, its velocity is 33i + 15j.
a) What was its velocity at the beginning of the 10s interval?
b) By how much did it's speed change?
c) By how much did its direction change?
d) Show that the speed change is not given by the magnitude of the acceleration multiplied by time. why?

The Attempt at a Solution

a) a→= Δv/Δt

∴
vi→ = (33i + 15j) - (23i + 36j)
vi→ = (10i -21j) ms^-1

b) |vi→| = 23.26ms^-1
|vf→| = 36.25ms^-1
speed = |v| = (|vi→|) -|vf→|
|v| = 13ms^-1

c)tanΘ = 15/33
Θ=24.45°

d) |a→| = sqrt (2.3^2 + 3.6^2) = 4.272ms^-1
Δt = 10s

|a→|.10s \neqΔv
∴4.272ms^-1 . 10s \neq13ms^-1

what is the explanation?

 

[hav0c]

what do you know about the initial directions of the velocity and acceleration?

 

[negation]

what do you know about the initial directions of the velocity and acceleration?

Could you be more specific?

 

[ehild]

How are velocity and speed defined? Is it possible that the velocity changes and the speed does not?
If the velocity is \vec {v } = \vec {v_0} +\vec {a } t what is the change of the speed during time t?

ehild

 

[negation]

How are velocity and speed defined? Is it possible that the velocity changes and the speed does not?
If the velocity is \vec {v } = \vec {v_0} +\vec {a } t what is the change of the speed during time t?

ehild

Is this question with respect to part (d)?

 

[ehild]

Is this question with respect to part (d)?

yes, of course.

ehild

 

[negation]

yes, of course.

ehild

Are my reasoning and workings valid from part(a) - part(c)?

In respond to your earlier question, the change in velocity is a function of time.
vf = vi + at

a.t = Δv
a = (2.3i + 3.6j)

Δv = 10*(2.3i + 3.6j) = 23i + 36j

 

[ehild]

Homework Statement

An object undergoes acceleration 2.3i + 3.6j for 10s. At the end of this time, its velocity is 33i + 15j.
a) What was its velocity at the beginning of the 10s interval?
b) By how much did it's speed change?
c) By how much did its direction change?
d) Show that the speed change is not given by the magnitude of the acceleration multiplied by time. why?

The Attempt at a Solution

a) a→= Δv/Δt

∴
vi→ = (33i + 15j) - (23i + 36j)
vi→ = (10i -21j) ms^-1

b) |vi→| = 23.26ms^-1
|vf→| = 36.25ms^-1
change of speed = Δ|v| = (|vi→|) -|vf→|
Δ|v| = 13ms^-1

It was the change of speed.

c)tanΘ = 15/33
Θ=24.45°

That is the angle of the final velocity Θf. What is the initial angle? By how much did the direction change?

d) |a→| = sqrt (2.3^2 + 3.6^2) = 4.272ms^-1
Δt = 10s

|a→|.10s \neqΔv
∴4.272ms^-1 . 10s \neq13ms^-1

what is the explanation?

The change of velocity is

\vec v_f - \vec v_i=\vec a t

You can make a triangle with side-lengths vf,vi and at. What do you know about the sides of a triangle? Can you draw a triangle with one side equal to the difference of the two other sides? How does it look like? What relation is true among the side lengths of a general triangle?

 

[negation]

It was the change of speed.


That is the angle of the final velocity Θf. What is the initial angle? By how much did the direction change?

final angle = 24.4 degrees
initial angle = arctan (-21/10) = -64.5 degrees
Θ_change = Θ_f - Θ_i = 89°

The change of velocity is

\vec v_f - \vec v_i=\vec a t

You can make a triangle with side-lengths vf,vi and at. What do you know about the sides of a triangle? Can you draw a triangle with one side equal to the difference of the two other sides? How does it look like? What relation is true among the side lengths of a general triangle?

But isn't the change in velocity mathematically reasoned to be a.t?

dv/dt = a
therefore,
a.dt = dv

Edit: it's the same thing.

In general, the sides can be expressed as ratio of one another. This property effective allows us to determine the angle between the lengths.

 

[ehild]

final angle = 24.4 degrees
initial angle = arctan (-21/10) = -64.5 degrees
Θ_change = Θ_f - Θ_i = 89°




But isn't the change in velocity mathematically reasoned to be a.t?

dv/dt = a
therefore,
a.dt = dv

Edit: it's the same thing.
View attachment 65773

In general, the sides can be expressed as ratio of one another. This property effective allows us to determine the angle between the lengths.

Yes, the change of velocity (vector) is the acceleration (vector) multiplied by time. But that is not true to the change of speed.
Look at your triangle. The length of one side is equal to the initial speed |v1|, the other is equal to the final speed |v2|, and the length of the third side is |a|t - magnitude of acceleration multiplied by time. Is |v2|-|v1|=|a|t?

ehild

 

[negation]

Yes, the change of velocity (vector) is the acceleration (vector) multiplied by time. But that is not true to the change of speed.
Look at your triangle. The length of one side is equal to the initial speed |v1|, the other is equal to the final speed |v2|, and the length of the third side is |a|t - magnitude of acceleration multiplied by time. Is |v2|-|v1|=|a|t?

ehild

Yes I noticed.

vf - vi = a.t \neq sqrt[vf^2 + vi^2]

Edit: yes it.

 

[ehild]

You need to write some explanation which holds for any vector difference. In this case, the initial and final velocities are almost perpendicular. Compare the magnitude of the difference and the difference of the magnitudes in general case.

ehild

 

[negation]

You need to write some explanation which holds for any vector difference. In this case, the initial and final velocities are almost perpendicular. Compare the magnitude of the difference and the difference of the magnitudes in general case.

ehild

I'll contemplate on the reasoning symbolically.