An object accelerates at a constant rate for a time t. Which equation gives its final velocity?
A car pulls away from traffic lights at 3 m/s and accelerates at 2 m/s² for 5 s. How fast is it moving at the end?
To solve this kind of problem, there are three steps to follow.
Step 1 — write down what we know, and mark what we need to find.
Step 2 — select the correct equation. Here, we can use: v = v₀ + at
Step 3 — substitute in the values, and calculate v.
Why does this equation work? a is the velocity gained each second. So in t seconds, the car gains a × t of velocity.
Add that gain to the velocity it started with, v₀, and you have the final velocity, v.
Further along the road, the driver brakes: from 13 m/s, the car's acceleration is −2 m/s² for 4 s. How fast is it moving then?
Write down the values in the question:
v = ?
v₀ = 13 m/s
a = −2 m/s²
t = 4 s
Write down the equation:
v = v₀ + at
Substitute in the values:
v = 13 + (−2)(4)
Simplify:
v = 13 − 8
v = 5 m/s
A truck moving at 25 m/s brakes with a constant acceleration of -3 m/s² for 5 s. Calculate its final velocity, in m/s.
A train moving at 12 m/s accelerates at a constant 2.5 m/s² for 6 s. Calculate its final velocity, in m/s.
Back at the light: it turns green, and five seconds later the car is doing 13 m/s — 3 m/s to start, 2 m/s more every second. v = v₀ + at said so before the wheels turned.
A cart on a straight track moves at -6 m/s — 6 m/s in the negative direction — and accelerates at a constant -2 m/s² for 4 s. Calculate its final velocity, in m/s. Include the sign in your answer.
A cyclist moving at 3 m/s accelerates at a constant 1.5 m/s² for 6 s. Calculate the cyclist's final velocity, in m/s.
A car moving at 9 m/s brakes with a constant acceleration of -2 m/s² for 3 s. Calculate its final velocity, in m/s.