Answer:
Given:
Initial speed (u) = 22 m/s
Final speed (v) = 0 m/s (Rest)
Time taken (t) = 4 seconds
To Find:
Distance travelled by car (s)
Explanation:
From equation of motion of object moving with uniform acceleration in straight line we have:
By substituting value of v, u & t in the equation we get:
Distance travelled by car (s) = 44 m
Answer:
≈ 6.68 m/s
Explanation:
A suitable formula is ...
vf^2 -vi^2 = 2ad
where vi and vf are the initial and final velocities, a is the acceleration, and d is the distance covered.
We note that if the initial launch direction is upward, the velocity of the ball when it comes back to its initial position is the same speed, but in the downward direction. Hence the problem is no different than if the ball were initially launched downward.
Then ...
vf = √(2ad +vi^2) = √(2·9.8 m/s^2·1.0 m+(5 m/s)^2) = √44.6 m/s
vf ≈ 6.68 m/s
The ball hits the ground with a speed of about 6.68 meters per second.
__
We assume the launch direction is either up or down.
Answer:
11 m/s
Explanation:
Draw a free body diagram. There are two forces acting on the car:
Weigh force mg pulling down
Normal force N pushing perpendicular to the incline
Sum the forces in the +y direction:
∑F = ma
N cos θ − mg = 0
N = mg / cos θ
Sum the forces in the radial (+x) direction:
∑F = ma
N sin θ = m v² / r
Substitute and solve for v:
(mg / cos θ) sin θ = m v² / r
g tan θ = v² / r
v = √(gr tan θ)
Plug in values:
v = √(9.8 m/s² × 48 m × tan 15°)
v = 11.2 m/s
Rounded to 2 significant figures, the maximum speed is 11 m/s.