Solve Kinematics: Find Velocity & Displacement of Speedboat

  • Thread starter Thread starter iaberrant
  • Start date Start date
  • Tags Tags
    Kinematics
AI Thread Summary
The speedboat accelerates in three stages: first at +2.03 m/s² for 8.87 seconds, then at +0.380 m/s² for 8.86 seconds, and finally decelerates at -1.20 m/s² for 8.11 seconds. The calculated velocity at 25.84 seconds is 7.28 m/s. The total displacement requires integrating the velocity over time, which can be visualized using a velocity-time diagram to find the area under the curve. Assistance is needed to calculate the total displacement accurately. Understanding the integration of velocity over time is crucial for solving the displacement part of the problem.
iaberrant
Messages
13
Reaction score
0

Homework Statement



A speedboat starts from rest and accelerates at +2.03 m/s2 for 8.87 s. At the end of this time, the boat continues for an additional 8.86 s with an acceleration of +0.380 m/s2. Following this, the boat accelerates at -1.20 m/s2 for 8.11 s. (a) What is the velocity of the boat at t = 25.84 s? (b) Find the total displacement of the boat.

Homework Equations


The equations to be used are :
v= u + at
s= ut + 1/2 at^2

The Attempt at a Solution



i understand that you use the above equations to work through each stage in the speedboats journey:
stage 1: v=11.69m/s
s=40.23m
stage 2: using the previous velocity
v=16.47m/s
s=102.236 m
stage 3: using the previous velocity
v= 7.28
s= 100.11

so the solution is velocity at 25.84 seconds is 7.28 m/s that part of my working is correct, what i don't understand is how to find the total displacement, please help
 
Physics news on Phys.org
Draw a v-t diagram and calculate a surface under the lines.
 
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
I treat this question as two cases of Doppler effect. (1) When the sound wave travels from bat to moth Speed of sound = 222 x 1.5 = 333 m/s Frequency received by moth: $$f_1=\frac{333+v}{333}\times 222$$ (2) When the sound wave is reflected from moth back to bat Frequency received by bat (moth as source and bat as observer): $$f_2=\frac{333}{333-v}\times f_1$$ $$230.3=\frac{333}{333-v}\times \frac{333+v}{333}\times 222$$ Solving this equation, I get ##v=6.1## m/s but the answer key is...
Back
Top