acceleration kinematics

Question

At a nearby university, Professor Jones teaches a course called “circus stunts 101”. In the first lab of the year, students are launched out of a cannon and into a safety net, hopefully. When launched, the students undergo an acceleration of 15.0 m/s² for 1.2 s before being hurled out of the cannon. (5 marks)

a. Find the length of the cannon.

b. Find the velocity of the student as she leaves the cannon.

Answer 1

Hi Stephen! Before we begin this problem, let's take a quick look at what information is provided to us by the problem and what information they have left for us to find. It seems that they have provided us with an acceleration and the time for which the object is accelerating. It can also be inferred that the object begins from rest, so we know our initial velocity as well. With this information, they want us to find the length of the cannon (a distance) and the final velocity of a student as she leaves the cannon.

Part a) To find the length of the cannon, we are essentially calculating the distance covered by the student as she accelerates for 1.2s. That means we can use the following equation to help us calculate that distance:

Δx = v₀t + (1/2)at²,

where Δx is the change in displacement, v₀ is the initial velocity, a is the acceleration, and t is time. Plugging in our initial velocity, acceleration, and time that we already know gives us

Δx = 0 * 1.2 + (1/2) * 15 * 1.2².

From here, we can solve for Δx, which is equal to the length of the cannon.

Δx = 0 + 10.8

Δx = 10.8

So, now we know that the length of the cannon is 10.8 m.

Part b) To calculate the final velocity of the student as she leaves the cannon, we can use the following equation:

Δv = v_f - v_i = at,

where Δv is the change in velocity, v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is time. Plugging in what we know, we get

v_f - 0 = 15 * 1.2.

Then, solving for v_f, we get

v_f = 15 * 1.2

v_f = 18.

So, we now know that the student's final velocity equals 18 m/s.