1. How would you expect an instantaneous acceleration vs. time graph to look for a cart moving with a constant
1 answer:
Answer:
a=0 v = v₀ + a t
a=0 line is horizontal
Explanation:
1, In a graph of acceleration vs. time, we have lines, when the line is horizontal it is zero, when the line has a positive slope the increasing accelerations and when the slope is negative the decreasing acceleration
2, speed and relationship of a car is given by
v = v₀ + a t
where vo is the initial velocity, a is the acceleration and tel time
in this case I will calcograph velocity vs. time the constant acceleration is a straight line.
In general from the graph we can find the initial velocity with the cut at that x and the acceleration of the car with the slope
You might be interested in
Answer:
Q1: 3.2km
Q2: 4.8K
Explanation:
Q1:
So db is the distance of bird, and dr is the distance of runner
db = 2vr and the distance of bird is going to be 2 times greater than the runner.
formulas: db = 2vr & db = 2dr
- db = 2dr
- L + (L - x) = 2x
- 2L - x = 2x
- 2L = 3x
- x =
L
Insert it in x = L
(2.4km) = 1.6km
Now we use formula db = 2dr
- db = 2L - x
- db = 2(2.4km) - 1.6km
- <u>db = 3.2km</u>
Q2:
Formulas: Vr = L /Δt & Vb = db/Δt
- Vr = L/ Δt ⇒ Δt =
(Km cancel each other)
- Vb = db/Δt ⇒ db = VbΔt
- 13.6km/hr
- <u>4.8km</u>
(hr cancel each other)
Hope it helps you :)