Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

# Physics Question: A vehicle accelerates from rest to 13.5 m/s in 12 seconds. How far did the vehicle travel? What formula(s) do you use?

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• MyRank
Lv 6
1 month ago

Given,

Initial speed (u) = 0m/sec

Final speed (v) = 13.5m/sec

Time (t) = 12sec

Distance (d) =?

Using kinematic relation:-

v = u + at

13.5m/sec = 0m/sec + a x 12

13.5m/sec = a x 12

Acceleration (a) = 13.5/12 = 1.125m/sec²

(13.5)² = (0)² + 2 x a x d

182.25 = 2 x 1.125 x d

Distance (d) = 182.25 / 2.25 = 81m

∴ Distance (d) = 81m.

• Anonymous
1 month ago

Vavg = Vfinal / 2 = 13.5m/s / 2 = 6.75 m/s

and then

distance d = Vavg * t = 6.75m/s * 12s = 81 m ◄

OR you could find the acceleration

a = Vfinal / t

and then

d = ½ * a * t²

with, of course, the same result.

Hope this helps!

• Anonymous
1 month ago

stretched distance d = (Vf+Vi)/2*t = (13.5+0)*12/2 = 13.5*6 = 27*3 = 81.0 m

or

acceleration a = ΔV/Δt = (13.5-0) / (12-0) = 27/24 = 9/8 of m/sec^2

stretched distance d = a/2*t^2 = 9/16*12^2 = 144*9/16 = 81.0 m

or

stretched distance d = (Vf^2-Vi^2) / 2a = (27^2/4-0)*8/18 = 27^2*2/18 = 81.0 m

• CRR
Lv 7
1 month ago

V=at

D=at^2/2

(Assuming constant acceleration)

You know V and t so you can find a and D.