When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.
Now, given that:
- the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.
- after hitting the wall, the final velocity will be (v) = 0 km/h
Assumptions:
- Suppose we make an assumption that the distance travelled during the collision of the car with the brick wall (S) = 1 m
- That the car's acceleration is also constant.
∴
For a motion under constant acceleration, we can apply the kinematic equation:
where;
v = final velocity
u = initial velocity
a = acceleration
s = distance
From the above equation, making acceleration (a) the subject of the formula:
The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.
since 1 km/h = 0.2778 m/s
100 km/h = 27.78 m/s
a = - 385.86 m/s²
Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration and velocity is;
v = u + at
where;
v = 0
-u = at
t = 0.07 seconds
An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.
Thus, we can conclude that the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.
Learn more about the kinematic equation here:
brainly.com/question/11298125?referrer=searchResults
It will help them stay in better shape resulting in better health, being more physically active, and staying happier as a result.
Answer:
The minimum distance in which the car will stop is
x=167.38m
Explanation:
∑F=m*a
∑F=u*m*g
The force of friction is the same value but in different direction of the force moving the car so it can stop so
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = 
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
