a) For the motion of car with uniform velocity we have , 
, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.
In this case s = 520 m, t = 223 seconds, a =0 
Substituting
The constant velocity of car a = 2.33 m/s
b) We have 
s = 520 m, t = 223 seconds, u =0 m/s
Substituting
Now we have v = u+at, where v is the final velocity
Substituting
v = 0+0.0209*223 = 4.66 m/s
So final velocity of car b = 4.66 m/s
c) Acceleration = 0.0209
Answer:
The forces that do non-zero work on the block are gravity and normal reaction force
Explanation:
Answer:
1.6 m
Explanation:
Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.
The time for landing should be calculated by using the second equation of motion formula
h = Ut + 1/2gt^2
Let U = 0
0.5 = 1/2 × 9.8 × t^2
0.5 = 4.9t^2
t^2 = 0.5 / 4.9
t^2 = 0.102
t = 0.32 s
The target should be placed so that the toy car lands on it at:
Distance = 5 × 0.32
distance = 1.597 m
Distance = 1.6 m
Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.
The working equation for this problem is written below:
x = v₀t + 0.5at², where x is the distance traveled, v₀ is the initial velocity, a is the acceleration and t is the time
Let's apply the concept of calculus. The maximum speed is equated to the derivative of x with respect to t.
dx/dt = 8 ft/s = v₀ + at
Since the it starts from rest, v₀ = 0
8 = at
t = 8/a
The net acceleration is 0.7 - 0.35 = 0.35 ft/s². Thus,
t = 8/0.35 = 22.86 seconds
