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PhysicsClassical Physics

Constant Acceleration Formula: Find a Dragster’s Final Velocity Without Time

Converting 0.250 mi to about 402 m and applying the no-time constant-acceleration equation gives a final dragster velocity of about 145 m/s.

Question

A dragster’s finish-line time is unknown. It accelerates from rest at 26 m/s² over a quarter mile (0.250 mi). Find its final velocity.

Answer

The textbook Solution first converts the quarter mile to meters:

(0.250 mi)(1609 m1 mi)≈402 m.(0.250\,\text{mi})\left(\frac{1609\,\text{m}}{1\,\text{mi}}\right)\approx402\,\text{m}.

The worked example then uses the constant-acceleration equation that eliminates time:

v2=v02+2aΔx.v^2=v_0^2+2a\Delta x.

The dragster starts from rest, so v0=0v_0=0. Substitute the known values:

v2=2(26.0 m/s2)(402 m)=2.09×104 m2/s2.v^2=2(26.0\,\mathrm{m/s^2})(402\,\mathrm m) =2.09\times10^4\,\mathrm{m^2/s^2}.

Taking the physically relevant positive square root,

v≈144.6 m/s≈145 m/s.v\approx144.6\,\mathrm{m/s}\approx145\,\mathrm{m/s}.

So the dragster’s final velocity is about 145 m/s145\,\mathrm{m/s} in the direction of motion.

The negative mathematical square root is not used because the stated motion and acceleration are both toward the finish line.

Evidence boundary

The original prompt gives a quarter mile (0.250 mi), not 402 m. The Answer follows the textbook Solution by converting 0.250 mi to about 402 m and applying the constant-acceleration kinematic relation used in that worked example. It does not separately model changing acceleration, reaction time, traction limits, or aerodynamic drag.

Sources

These references support the concepts and methods used in the explanation above.

Constant Acceleration Formula: Dragster Velocity | Verla