The problem involves a dog accelerating from an initial velocity of 2.0 m/sec to a final velocity of 5.0 m/sec while covering a distance of 14.0 m. The goal is to determine the time taken for this acceleration, considering the equations of motion and the concept of average velocity.
Some participants have provided guidance on the correct interpretation of the formula and the need to consider acceleration. Multiple interpretations of the equation are being explored, with some participants suggesting the use of average velocity while others emphasize the need to calculate acceleration first.
There is an assumption of constant acceleration, and participants are navigating the implications of this assumption on the calculations. Some confusion exists regarding the correct formulation of the equations and the use of parentheses in mathematical expressions.
RachaelA said:d=(Vi+Vf/2)t
RachaelA said:Homework Statement
A dog accelerates from 2.0 m/sec to 5.0 m/sec while covering a distance of 14.0 m. How long did this take
Homework Equations
d=(Vi+Vf/2)t
The Attempt at a Solution
2(-Vi-Vf)d=t
I am doing something wrong. Please help. Thanks[/B]
This is the answer according to my teachervktsn0303 said:Sorry. A small correction in my previous reply.
t is not velocity*distance.
Seems like RachaelA has been taught to use the average velocity, being merely the average of initial and final velocities in the case of uniform acceleration. So no need to calculate the acceleration as such.Maged Saeed said:You have to put the acceleration in consideration ,
Please use parentheses correctly. What you have written means ##d = \left(V_i+\frac{V_f}{2}\right)t##, but I hope you mean ##d = \left(\frac{V_i+V_f}{2}\right)t##, which you could have written as d=((Vi+Vf)/2)t.RachaelA said:d=(Vi+Vf/2)t
Yes.RachaelA said:I think I should divide it?
Again, I hope you mean ##t = \frac d{\left(\frac{V_i+V_f}{2}\right)}## (which can be simplified a little, but no matter).RachaelA said:This is the answer according to my teacher
t=d/(Vf+Vi/2)