find the dimensions of the acceleration

Apr 2015
1,227
356
Somerset, England
Well b1 is not an acceleration and you are asked to find the dimensions of b1.

Do you know what dimensions in general are and particularly what the dimensions of acceleration are?
 
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Jan 2020
4
0
algeria
the dimensions of an acceleration are its components throught the y axis and the x axis
can you please help me solve this
 
Apr 2015
1,227
356
Somerset, England
the dimensions of an acceleration are its components throught the y axis and the x axis
can you please help me solve this
Absolutely not.

This was why i asked if you knew about dimensions.

You are right that the x and y values are the components of the vector acceleration.

But the dimensions are quite a different thing entirely.

Dimensions are connected to the units we use for a quantity and it often helps to write down the units when we want to find out the dimensions.

Dimensions are a way of relating the physical interpretation of any quantity to a few basic (fundamental) quantities.

For mechanics we have Mass, Length, and Time, symbols M, L and T.

(We add add a few more additioanl ones for thermal, electrical and optical quantities)

The dimension of a quantity is the product (not sum) of one or more of these basic quantites, each raised to a specific power.

This is best shown by examples.

The dimensions of distance are Length raised to the power 1, time raised to the power 0, and mass raised to the power 0


\(\displaystyle DimensionsofLength = {L^1}{T^0}{M^0} = L\)

velocity is


\(\displaystyle Velocity = \frac{{dis\tan ce}}{{time}} = {L^1}{T^{ - 1}}{M^0} = {L^1}{T^{ - 1}}\)

So if velocity is meteres per second than acceleration is metres per second per second can you follow this pattern and find the dimensions of acceleration for practice ?

Then we can look at finding the dimensions of B
 
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Jan 2020
4
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algeria
\(\displaystyle acceleration=velocity÷ time==L1 T−1 T1 = L1 T−2 \)
also sorry for not typing correctly iam still trying to figure this out
thank you so much i think i got it
that makes b=L1 T−2 T2 M0 =L1 T−4
if this is correct can you please help me solve the second question thanks in advance
 
Apr 2015
1,227
356
Somerset, England
\(\displaystyle acceleration=velocity÷ time==L1 T−1 T1 = L1 T−2 \)
also sorry for not typing correctly iam still trying to figure this out
thank you so much i think i got it
that makes b=L1 T−2 T2 M0 =L1 T−4
if this is correct can you please help me solve the second question thanks in advance
Yup you got it, are you sure you haven't met dimensions before?

As to the second part, this section is University physics, so i assume you can do simple integrations.

I also assume you can do the high school work on the parabola that the ball describes as this is under constant acceleration (gravity).

So let the horizontal distance travelled by the ball = h.
The the cyclist travels a distance = d + h.
They both travel for the same time t1.
From the equations for the parabola you can determine h and t1, since you are given vy0.

If you take the given equation between b the (instantaneous) acceleration and time you can integrate it to get a second equation.
This equation gives the the instantaneous velocity as a function of b and time, noting the value of v at t= 0 give the constant of integration.
Integrating this second equation gives a third equation connecting the distance travelled, time and b.

So you can now use this information to eliminate t1 (as required) and the distance travelled and report in terms of the variables required.
 
Jan 2019
100
75
So let the horizontal distance travelled by the ball = h.
The the cyclist travels a distance = d + h
If I'm reading the problem statement correctly, the ball is thrown "vertically upward" at position x(0) = d, while the position of the bicycle at that same time is x(0) = 0.

The ball has no horizontal velocity.

The time required for the ball to go up and down is the same as the time it takes the bicycle to travel from x=0 to x=d, so that the bicyclist may catch the ball at the height that it was launched upward.