physics question

Question

The friction of the water on a boat produces an acceleration of -10. m/s². If the boat is traveling at 30. m/s and the motor is shut off, how long does it take the boat to slow down to 5.0 m/s? Show me with equations like delta x etc.

Answer 1

Write out all of your known information, and, if you can, draw a picture of what's happening:

1. Initial velocity, v_i = 30 m/s
2. Final velocity, v_f = 5 m/s
3. Acceleration due to friction, a_fric = -10 m/s²
4. We are solving for time, t
5. We need the equation for velocity, which is as follows:
6. v_f = v_{i +} a_{fric *} t
7. In words, this equation says "the final velocity of an object is equal to whatever starting velocity it had plus its acceleration rate times however long it was accelerating"
8. v_f - v_i can also be represented as Δv, which will come in handy later. Delta Δ denotes a change in value. In this case, the change in velocity over some given amount of time

You can either substitute values now and solve for t, or you can re-write the equation isolating the variable t and then substituting values after. I will show both.

1. Substituting values first, we have:
2. 5 m/s = 30 m/s + (-10 m/s²) * t
3. Subtract v_i from both sides (30 m/s)
4. -25 m/s = -10 m/s² * t
5. Divide both sides by the acceleration a_fric (-10 m/s²) to find the answer
6. -25 m/s ÷ -10 m/s² = t
7. t = 2.5 seconds

Now, rearranging the equation first to isolate t and then substituting values will give us the same answer, but this may help you better understand the underlying math at work.

1. First, write out the equation
2. v_f = v_{i +} a_{fric *} t
3. Subtract the initial velocity from both sides
4. v_f - v_i = a_{fric *} t
5. v_f - v_i = Δv
6. Divide the change in velocity, Δv, by the rate of acceleration to isolate the variable t
7. Δv ÷ a_{fric =} t
8. Plug in numbers
9. Δv = 5 m/s - 30 m/s
10. Pay attention to the initial and ending velocities, Δv is v_f - v_i NOT v_i - v_f
11. Δv = -25 m/s
12. -25 m/s ÷ -10 m/s² = 2.5 seconds

A note on dimensional analysis: when dividing a velocity by an acceleration, the meters per second will cancel out, leaving (1/(1/s)) for units, which is the same as 1s. I can explain this more clearly in a tutoring session if you need extra help here.

Answer 2

the problem asking for time t.

Using formular: v = vo + at

where v = 5 m/s, vo = 30m/s, a = -10m/s^2

to get t = 2.5s

Bonus, if it as for how far the boat has gone to reach the velocity of 5m/s,

You can try this formular: v^2 - vo^2 = 2 a.s

where: v: the velocity after (5m/s)

vo: velocity before ( 30m/s)

a: acceleration (-10m/s)

s: the distance travelled when velocity is v to vo

base on the formular, we get distance is 43.75m.