Answer:
1) Time interval Blue Car Red Car
0 - 2 s Constant Velocity Increasing Velocity
2 - 3 s Constant Velocity Constant Velocity
3 - 5 s Constant Velocity Increasing Velocity
5 - 6 s Constant Velocity Decreasing Velocity
2) For Red and Blue car y₂ = 120 v = 
= 
= 20 m/s
We get the same velocity for two cars because it is the average velocity of the car at the given interval of time. It is measured for initial and final position.
3) At t = 2s, the cars are the same position, and are moving at the same rate
Position - same
Velocity - same
The position-time graph shares the same spot for two cars.
Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.
Let's break them down into components.
X Y
v₁ 32 cos50 m/s 32 sin50 m/s
v₂ 32 cos50 m/s ?
Δd ? 0
Δt ? ?
a 0 -9.8 m/s²
Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.
Δdy = v₁yΔt + 0.5ay(Δt)²
0 = v₁yΔt + 0.5ay(Δt)²
0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0
0 = v₁ + 0.5ayΔt
0 = 32sin50m/s + 0.5(-9.8m/s²)Δt
0 = 2<u>4</u>.513 m/s - 4.9m/s²Δt
-2<u>4</u>.513m/s = -4.9m/s²Δt
-2<u>4</u>.513m/s ÷ 4.9m/s² = Δt
<u>5</u>.00s = Δt
Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.
Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²
Δdₓ = 32cos50m/s(<u>5</u>.00s) + 0.5(0)(<u>5</u>.00)²
Δdₓ = 32cos50m/s(<u>5</u>.00s)
Δdₓ = 10<u>2</u>.846
Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.
I'm not entirely sure, but I think the first is A, and the second is inverted.
Answer:
1. Velocity = 75.39 m/s
2. Air resistance
3. Mass of the coin
Explanation:
1. Given that the height of the building is 290 m. From Newton's third law of motion,
= 
+ 2gh
where: V is the final velocity of the coin, U is the initial velocity, g is the force of gravity and h is the height.
Since the coin was dropped from a height, its initial velocity is zero. The force of gravity is taken as 9.8 m/
, so that:
= 0 + 2 x 9.8 x 290
= 5684
V = 
= 75.3923
The velocity with which the coin hits the ground is 75.39 m/s.
2. Air resistance: During the free fall of the coin, the direction of wind flow could either cause an increase or decrease the value predicted.
ii. Mass of the coin: This can also affect the predicted value.
Part a)
here as per the situation shown we can say that the tension force at the position of applied force is given as
here
= angle of rope with horizontal
so we will have
now we can say
so the tension force in the string is 1855.2 N
Part B)
this method is better then pulling the car directly because if we pull the car directly then the force on the car is 610 N while in this case we get almost 3 times more force so it is better way to pull the car
