Solve Boat, Seagull & Speeder Problems: Acceleration, Speed & Time

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In summary: Once you have that, you can use the equations of motion to solve for the remaining variables, such as acceleration, maximum height, and time of flight.
  • #1
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1)A boat is cruising in a straight line at a constant speed of 2.0 m/s when it is shifted into neutral. After coasting 10.0 m the engine is engaged again, and the boat resumes cruising at the reduced speed of 1.6 m/s. Assume constant acceleration while coasting.
(a) How long did it take for the boat to coast the 10.0 m?
(b) What was the boat's acceleration while it was coasting?

2) A seagull, ascending straight upward at 5.55 m/s, drops a shell when it is 10.5 m above the ground.
(a) What is the magnitude and direction of the shell's acceleration just after it is released?
(b) Find the maximum height above the ground reached by the shell.
(c) How long does it take for the shell to reach the ground?
(d) What is the speed of the shell at this time?

3) Referring to the Interactive Figure below, the speeder passes the position of the police car with a constant speed of 11.3 m/s. The police car immediately starts from rest and pursues the speeder with constant acceleration. What acceleration must the police car have if it is to catch the speeder in 12.6 s? Measure time from the moment the police car starts.

i am just not sure what formulas to use
any help would be awesome
thank you
 
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  • #2
You must use the equations of motion of course. They are

[tex]v^2 - u^2 = 2as[/tex]
[tex]v = u + at[/tex]
[tex]s = ut + \frac{1}{2}at^2[/tex]

Here, for any motion of constant acceleration

v = final velocity
u = initial velocity
s = displacement
a = constant acceleration
t = time

The above equations are all vector equations, i.e. all variables except time are vectors.

Can you identify your variables from here?
 
  • #3
thanks praharmitra
solved 3
still can't figure out 1 and 2
can you start me out on these two questions?
thanks
 
  • #4
for q1. we are basically concerned with the 10m that the boat is coasting. At what velocity does the boat start coasting and at what velocity does it stop coasting ?

How much distance does it travel while coasting?

Got three of your variables??

Now decide which equation to use...For q2: answer this: What is the velocity of the shell when it is dropped? remember the bird had some initial velocity.
 

Related to Solve Boat, Seagull & Speeder Problems: Acceleration, Speed & Time

1. What is acceleration?

Acceleration is the rate at which an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction. It is typically measured in meters per second squared (m/s^2).

2. How do you calculate acceleration?

Acceleration can be calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What is the difference between speed and velocity?

Speed is a scalar quantity that measures the rate at which an object is moving. It does not take into account direction. Velocity, on the other hand, is a vector quantity that measures the rate at which an object is moving in a specific direction.

4. How do you calculate speed?

Speed can be calculated by dividing the distance traveled by the time it took to travel that distance. The formula for speed is s = d/t, where s is speed, d is distance, and t is time.

5. How can acceleration, speed, and time be applied to solve boat, seagull, and speeder problems?

Acceleration, speed, and time can be used to solve problems involving the motion of boats, seagulls, and speeders by using the equations mentioned above. For example, if you know the acceleration of a boat and the time it takes to reach a certain speed, you can calculate the distance it traveled. Similarly, if you know the speed of a seagull and how long it flew, you can calculate the distance it covered. These equations help us understand and predict the motion of objects in real-life scenarios.

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