Car A and car B each gain 12 m/s of velocity. Car A takes 6 s; car B takes 3 s. Which car has the greater acceleration?
One car gains 8 m/s of velocity in 4 s. A rival gains 5 m/s in 2 s. One changes velocity by more; the other takes less time. Which acceleration is greater? This is a comparison you can't settle by eye. You need a number.
To decide whose acceleration is greater, acceleration needs a number of its own. This lesson gives acceleration its equation and its unit.
You've already seen Δ in Δx, the change in position. Δ on its own still means 'change in', so Δv reads 'the change in velocity'. Δv and Δt are each one whole symbol, read 'delta v' and 'delta t'.
The unit comes straight out of the division: (m/s) ÷ s gives m/s². Multiply instead and you would get (m/s) × s, which is m, a distance. Divide the wrong way up and you would get s²/m, an upside-down unit.
During a 40 s car ride, the car's velocity changes by 14 m/s over one 4 s stretch of track. To find the acceleration on that stretch with a = Δv/Δt, which value belongs in place of Δt?
Now the opening tie settles. Divide each car's change in velocity by the time it takes.
The first car gains 8 m/s of velocity in 4 s. Its rival gains 5 m/s in 2 s. Whose acceleration is greater?
Divide each car's change in velocity by the time it takes:
first car: a = 8/4 = 2 m/s²
rival: a = 5/2 = 2.5 m/s²
The rival's acceleration is greater: 2.5 m/s² against 2 m/s². The number settles what comparing by eye could not.
A boat gains 10 m/s of velocity over 5 s. What is its acceleration?
Write down the values in the question:
a = ?
Δv = 10 m/s
Δt = 5 s
Write down the equation:
a = Δv/Δt
Substitute in the values:
a = 10 / 5
Simplify:
a = 2 m/s²
A sprinter's velocity changes by 6 m/s during a 2 s burst. Which value belongs in place of Δv in a = Δv/Δt?
An airplane's velocity changes by 24 m/s during an 8 s takeoff run. Which substitution into a = Δv/Δt is correct?
Which statement says in words what a = Δv/Δt means?