On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C
Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)
on the DT graph im going to assume that movement B is a positive velocity with A and C being negative.
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity.
Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question.
i'll skip my working out as its the same as before:
C, B, A and then D
the same idea as on Q30 applies to Q31 part b,
D,C,B then A
Answer:
t = 2 v₀ / g
Explanation:
For this projectile launch exercise we use the displacement equations
x = vox t
y = y₀ +
t - ½ g t²
As it is launched horizontally the vertical velocity is zero and the point of origin of the coordinate system is here, so y₀ is zero.
x = v₀ t
y = ½ g t²
They ask us for the time for which
x = y
vo t = ½ g t²
t = 2 v₀ / g
Answer:
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