# Dimensional Analysis (University Physics)?

Can someone please answer and explain the procedure gone about to prove or derive answers for the following two problems

1) The acceleration (a) of an object is given by the equation a=At=Bt^2 , where t refers to time. what are the dimensions of A and B?

2) The speed, v, of an object is given by the equation v=At^3-Bt, where t is time. What are the dimensions of A and B?

Note: I was told making A and B the subject of the formula was the wrong approach as well as simultaneous equations

### 2 Answers

- oldprofLv 76 years agoBest Answer
1. a ~ m/s^2; so AT ~ m/s^2 and BT^2 ~ m/s^2 also.

So we have AT ~ As ~ m/s^2 and A ~ m/s^3. ANS. And BT^2 ~ Bs^2 ~ m/s^2 and B ~ m/s^4. ANS.

You should learn that units can be operated on just like variables: add, subtract, multiply and divide. In both these cases, we divided the time units.

2. v ~ m/s; so At^3 ~ m/s and As^3 ~ m/s so that A ~ m/s^4; and Bs ~ m/s; so B ~ m/s^2. ANS.

Note each of the two terms must have the same units as the LHS of the equation, the dependent variable v.

- RobertLv 46 years ago
acceleration has units m*s^-2.

the units of time is seconds. so something times seconds has the units of acceleration. What are these units?

Now repeat for B and 2).