Acceleration of a Dragster

The equation $v^2=v_0^2+2a\left(d-d_0\right)$ is ideally suited to this task because it gives the velocity from acceleration and displacement, without involving the time.

1. Convert miles to meters.
   $$(0.250\text{ mi})\times \frac{1609\text{ m}}{1\text{ mi}}=402\text{ m}$$

   3.9
2. Identify the known values. We know that v₀ = 0 since the dragster starts from rest, and we know that the distance traveled, d − d₀ is 402 m. Finally, the acceleration is constant at a = 26.0 m/s².
3. Insert the knowns into the equation $v^2=v_0^2+2a\left(d-d_0\right)$ and solve for v.
   $$v^2=0+2\left(26.0\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)(402\text{ m})=2.09\times {10}^4\frac{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{2}}}$$

   3.10

Taking the square root gives us $v=\sqrt{2.09\times {10}^4\frac{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{2}}}}=145\frac{\text{m}}{\text{s}}\text{.}$