Is it a ten mark question? Got to be more to it than speed = distance / time.
erm, dont you differentiate?
Question is what is velocity AT time t=whatever. NOT what is the average velocity after time t=x, which is what the other guys have done.
v(t) = 4t + 4
e: mufc, velocity is m/s not m/s/s
velocity = rate of change of position, differentiate and solve
4t+4 when t=1.8
11.2?
velocity = rate of change of position, differentiate and solve
4t+4 when t=1.8
11.2?
s=2t^2+4t-3 ... t=1.8
S = Distance
T = Time
V = Velocity
s=2(1.8)^2+4(1.8)-3 ... s=10.68m
s=vt Distance=Velocity*Time (this is a common formula you should know)
v=s/t
10.68/1.8 = 5.93ms^-2
This assumes a constant velocity which is not correct in this question. As others have stated, you need to differentiate to find the velocity.
's' is actually the "standard" letter for distance (where you're not wanting to imply distance in the horizontal direction).I think using the letter 's' to describe postion and not speed threw a few people off.
Examiners just looove to be sneaky like that![]()
's' is actually the "standard" letter for distance (where you're not wanting to imply distance in the horizontal direction).
("standard" in quotes because it's not set in stone. That is, it's a lot less standard than t for time or v for velocity).