Answer:
50m; 0m/s.
Explanation:
Given the following data;
Initial velocity = 20m/s
Acceleration, a = - 4m/s²
Time, t = 5secs
To find the displacement, we would use the second equation of motion;
Substituting into the equation, we have;
S = 50m
Next, to find the final velocity, we would use the third equation of motion;
Where;
- V represents the final velocity measured in meter per seconds.
- U represents the initial velocity measured in meter per seconds.
- a represents acceleration measured in meters per seconds square.
<em>Substituting into the equation, we have;</em>
V = 0m/s
<em>Therefore, the displacement of the bus is 50m and its final velocity is 0m/s.</em>
Consider the motion of the car before brakes are applied:
v₀ = maximum initial velocity of the car before the brakes are applied
t = reaction time = 0.50 s
x₀ = distance traveled by the car before brakes are applied
since car moves at constant speed before brakes are applied
Using the equation
x₀ = v₀ t
x₀ = v₀ (0.50)
Consider the motion after brakes are applied :
v₀ = initial velocity of the car before the brakes are applied
a = acceleration = - 10 m/s²
v = final velocity of the car after it comes to stop = 0 m/s
x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀ (0.50)
Using the equation
v² = v²₀ + 2 a x
inserting the values
0² = v²₀ + 2 (- 10) (38 - v₀ (0.50))
v²₀ = 20 (38 - v₀ (0.50))
v₀ = 23 m/s
Answer:
h = 20 m
Explanation:
given.
height, h = 10 m
Potential energy at 10 m = 50 J
Kinetic energy at 10 m = 50 J
maximum height the ball will reach, H = ?
Total energy of the system
T E = 50 J + 50 J
T E = 100 J
now,
A h = 10 m
P E = m g h
50 = m g x 10
mg = 5 ..............(1)
at the top most Point the only Potential energy will be acting on the body.
now, TE = Potential energy
100 = m g h
5 h = 100
h = 20 m
hence, the maximum height reached by the ball is equal to 20 m.
Ep= mgh
Ep = 40 x 9.8 x 10
Ep = 3920J
Ep = 3900J (2sf)
