Math  /  Algebra

Question6. Ben designed a human-powered submarine that can achieve a maximum speed of 3.06 m/s3.06 \mathrm{~m} / \mathrm{s}. If the submarine starts from rest and accelerates at 0.800 m/s20.800 \mathrm{~m} / \mathrm{s}^{2} until it reaches its maximum speed, then travels at this speed for another 5.00 s , what is the total distance traveled by the submarine?
Given

Studdy Solution

STEP 1

What is this asking? How far does Ben's submarine go if it speeds up from zero to its max speed, and then keeps going at that speed for a bit longer? Watch out! We've got two parts to this journey: the speeding-up part and the constant-speed part.
Don't mix them up!

STEP 2

1. Calculate the distance during acceleration.
2. Calculate the distance at constant speed.
3. Calculate the total distance.

STEP 3

Alright, so first, we need to figure out how far the submarine travels while it's accelerating.
We know the **initial speed** is 0 m/s0 \text{ m/s} (because it starts from rest), the **acceleration** is 0.800 m/s20.800 \text{ m/s}^2, and the **final speed** is 3.06 m/s3.06 \text{ m/s}.

STEP 4

We can use this super handy equation to find the distance traveled during acceleration: vf2=vi2+2adv_f^2 = v_i^2 + 2 \cdot a \cdot d.
Here, vfv_f is the **final velocity**, viv_i is the **initial velocity**, aa is the **acceleration**, and dd is the **distance**.

STEP 5

Let's plug in our values: (3.06 m/s)2=(0 m/s)2+2(0.800 m/s2)d(3.06 \text{ m/s})^2 = (0 \text{ m/s})^2 + 2 \cdot (0.800 \text{ m/s}^2) \cdot d.

STEP 6

Simplifying, we get 9.3636 m2/s2=1.6 m/s2d9.3636 \text{ m}^2\text{/s}^2 = 1.6 \text{ m/s}^2 \cdot d.

STEP 7

Now, **divide both sides** by 1.6 m/s21.6 \text{ m/s}^2 to solve for dd: d=9.3636 m2/s21.6 m/s2=5.85225 md = \frac{9.3636 \text{ m}^2\text{/s}^2}{1.6 \text{ m/s}^2} = 5.85225 \text{ m}.
So, the submarine travels about 5.85 m\textbf{5.85 m} while accelerating.

STEP 8

Next, we need to figure out how far it goes while traveling at its **maximum speed** of 3.06 m/s3.06 \text{ m/s} for 5.00 s5.00 \text{ s}.

STEP 9

For this, we can use the simple formula: distance=speedtime\text{distance} = \text{speed} \cdot \text{time}.

STEP 10

Plugging in the values, we get distance=(3.06 m/s)(5.00 s)=15.3 m\text{distance} = (3.06 \text{ m/s}) \cdot (5.00 \text{ s}) = 15.3 \text{ m}.

STEP 11

Finally, to find the **total distance**, we just **add** the distance traveled during acceleration and the distance traveled at constant speed.

STEP 12

Total distance =5.85 m+15.3 m=21.15 m= 5.85 \text{ m} + 15.3 \text{ m} = 21.15 \text{ m}.

STEP 13

The submarine travels a total distance of 21.15 m\textbf{21.15 m}.

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