I am trying to calculate the car's average acceleration in (m/s^2) between 0 and 2.1 s. I need to express this answer using two significant figures, I have been trying the following to get it and it keeps telling me my answer is wrong, what am I doing wrong?

a = (delta v)/(delta time)
a = (60 – 0) / (2.1)
a = 28.57

The gear ratios are gearing down, i.e. they reduce the angular velocity at the wheels and increase the torque. The correct formula is:

$$ T_{wheels} = T_{engine} G $$

Plus don't forget that there is a differential gear as well as the gearbox, and this reduces the angular velocity and increases the torque as well. The value of $G$ in your equation is the gearbox ratio multiplied by the differential ratio.

Response to comment:

I don't know what the differential gearing is in your car, but differentials are generally around 4:1 so let's take this value. In that case the torque at the wheels in first gear is:

The acceleration of an object at a given time $t$ is, in fact, defined as the average acceleration of that object for a small interval around that time in the limit that the interval becomes "infinitely small." The mathematical concept of the limit makes the notion of "infinitely small" precise.

Suppose $v(t)$ represents the velocity of the object moving along a straight line at time $t$. If we want to find its acceleration at that time, what we do is to first determine its average acceleration for some interval of time between time $t$ and $t+\Delta t$ where $\Delta t >0$. This is just given by the change in velocity divided by the change in time over that interval;
$$
a_\mathrm{average}(t,t+\Delta t) = \frac{v(t+\Delta t) - v(t)}{\Delta t}
$$
Next, we imagine taking that interval infinitely small; the result is the instantaneous acceleration at time $t$. This operation is made notationally precise by the following mathematical notation, and the precise mathematical notion can be defined in terms of the "delta-epsilon" definition found in standard calculus and real analysis texts:
$$
a(t) = \lim_{\Delta t \to 0} a_\mathrm{average}(t, t+\Delta t).
$$
Putting our two equations together gives the definition of instantaneous acceleration:
$$
a(t) = \lim_{\Delta t\to 0} \frac{v(t+\Delta t)-v(t)}{\Delta t}
$$
The right hand side of this expression is mathematically referred to as the derivative of the velocity function and is often denoted in physics with an overdot;
$$
\dot v(t) = \lim_{\Delta t\to 0} \frac{v(t+\Delta t)-v(t)}{\Delta t}.
$$
This allows us to restate the definition of acceleration succinctly as
$$
a(t) = \dot v(t)
$$
Or said in words, acceleration is the time derivative of velocity.

## Best Answer

What is wrong is the units. You are doing

$$ \frac{(60 \mbox{ miles/hr})}{(2.1 \mbox{ sec})} = 28.57 \mbox{ mph/s } $$

What you need to do is convert the speed into $\mbox{m/s}$

$$ (60 \mbox{ miles/hr}) = ( \frac{1}{3600}60 \mbox{ miles/sec} ) = ( 1610 \frac{1}{3600} 60 \mbox{ meter/sec}) = 26.833 \mbox{ m/s} $$

Now do the acceleration as

$$ \frac{(26.833 \mbox{ meter/sec})}{(2.1 \mbox{ sec})} = ... \mbox{m/s}^2 $$