Dragsters can achieve average accelerations of 25.0m/s2. Suppose such a dragster accelerates from rest at this rate for 5.36s. How far does the dragster travel in this time?

Answer 1

Take Distance Traveled as Average Velocity Times Time Traveled. Write this as x = v_averaget or

(v₀ + v)t/2.

Next consider Velocity (v) as Initial Velocity (v₀) Plus Acceleration Times Time (at) or v₀ + at.

Then x equal to v_averaget or (v₀ + v)t/2 becomes (v₀ + v₀ + at)t/2 which goes to 0.5(2v₀t + at²)

or v₀t + 0.5at²

For this problem v₀ = 0 meters per second, a = 25 meters per second squared, and t = 5.36 seconds.

Then build x = (0 m/s)(5.36 s) + 0.5(25 m/s²)(5.36 s)² which simplifies

to 359.12 meters.

Note that methods of Calculus enable one to state:

Displacement With Respect To Time (That Is, Distance) Equals v₀t + 0.5at² ;

Instantaneous Rate Of Change In Distance With Respect To Time (Or Velocity)

Is Given By dx/dt Equal To d(v₀t)/dt + 0.5d(at²)/dt or v₀ + 0.5(2at^(2-1)) or v₀ + at ;

Instantaneous Rate Of Change In Velocity With Respect To Time (Or Acceleration)

Is Given By d²x/dt² = dv₀/dt + d(at)/dt or 0 + a or a.