Instantaneous Acceleration Given Equation for Velocity

[amandela]

This is from an old exam.

The velocity of a particle moving along a straight line is v = 4 + 0.5 t. What is the instantaneous acceleration at t=2?

The solution is supposedly 2 because a = dv/dt = t. But I thought dv/dt here would be 0.5. What am I missing?

Thanks.

 

[berkeman]

Welcome to PF.

Where are you seeing a solution of 2? dv/dt for that equation is indeed 0.5 if you've written the equation correctly.

 

[Orodruin]

v = 4 + 0.5 t

Should this be $v= 4 + 0.5 t^2$??

 

[amandela]

Welcome to PF.

Where are you seeing a solution of 2? dv/dt for that equation is indeed 0.5 if you've written the equation correctly.

Thank you. I think I lost an exponent!

 

[amandela]

Should this be $v= 4 + 0.5 t^2$??

Yes, I went back and checked again b/c the displacement was wrong, too, and I missed the exponent. Thank you!

 

[PeroK]

This is from an old exam.

The velocity of a particle moving along a straight line is v = 4 + 0.5 t^2

What's your opinion about the dimensionality of that equation? The lefthand side is a velocity and the righthand side is a number plus a time squared. Is that valid?

 

[jbriggs444]

What's your opinion about the dimensionality of that equation? The lefthand side is a velocity and the righthand side is a number plus a time squared. Is that valid?

In my opinion the equation is only valid for a set of units where the appropriate conversion constants all turn out to be equal to one.

There should be units on the $4$ and units on the $0.5$. The computed result would then have units for the $v$. Also units for the given value of $t = 2$. The first derivative of that would then have units for the acceleration.

 

[Orodruin]

What's your opinion about the dimensionality of that equation? The lefthand side is a velocity and the righthand side is a number plus a time squared. Is that valid?

Lower level textbooks often introduce dimensionless quantities such as ”velocity $v$ m/s or simply do not use units at all.