Which step should you take before you begin a scientific investigation? Print the directions after you complete the experiment.
2 answers:
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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:
Answer:
The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s
Explanation:
The given parameters of the ball are;
The initial speed of the ball = 15 m/s
The direction in which the ball is launched = 50° above the horizontal
The location of the other tennis player when the ball is launched = 10 m from the ball
The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched
The height at which the ball is hit back = 2.1 m above the height from which the ball is launched
The vertical position, 'y', at time, 't', of a projectile motion is given as follows;
y = (u·sinθ)·t - 1/2·g·t²
When y = 2.1 m, we have;
2.1 = (15·sin(50°))·t - 1/2·9.8·t²
∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0
Solving with the aid of a graphing calculator function, we get;
t = 0.199776187257 s or t = 2.14525782198 s
Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds
The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;
x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m
The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m
Therefore, we have;
The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s
The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m
The minimum average speed the other player has to move with, = d/t₂
∴ = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s
The minimum average speed the opponent must move so that he is in position to hit the ball, ≈ 5.79 m/s.