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6. a skateboarder is moving at 1 m/s and accelerates at a rate of 0.1 m…

Question

  1. a skateboarder is moving at 1 m/s and accelerates at a rate of 0.1 m/s² over a distance of 25 meters. how fast are they going now?
  2. if a bird is flying at 4.0 m/s and accelerates at 1 m/s² for 3 seconds, how far would it fly in that time?
  3. if you are riding a bike at 5 m/s and hit the brakes and decelerate at 0.5 m/s² for 3 seconds, how fast would you be going?
  4. if you start from rest (not moving) and accelerate at 0.25 m/s² for 5 seconds, how fast would you be going?
  5. if you start from rest and accelerate at 0.3 m/s² for 4 seconds, how far would you go in that time?

Explanation:

Response
Question 6

Step1: Identify the kinematic equation

We use the equation \( v^2 = u^2 + 2as \), where \( u = 1 \, \text{m/s} \), \( a = 0.1 \, \text{m/s}^2 \), \( s = 25 \, \text{m} \).

Step2: Substitute the values

\( v^2 = (1)^2 + 2\times0.1\times25 \)
\( v^2 = 1 + 5 \)
\( v^2 = 6 \)

Step3: Solve for \( v \)

\( v = \sqrt{6} \approx 2.45 \, \text{m/s} \)

Step1: Identify the kinematic equation

We use the equation \( s = ut + \frac{1}{2}at^2 \), where \( u = 4.0 \, \text{m/s} \), \( a = 1 \, \text{m/s}^2 \), \( t = 3 \, \text{s} \).

Step2: Substitute the values

\( s = 4.0\times3 + \frac{1}{2}\times1\times(3)^2 \)
\( s = 12 + 4.5 \)
\( s = 16.5 \, \text{m} \)

Step1: Identify the kinematic equation

We use the equation \( v = u + at \), where \( u = 5 \, \text{m/s} \), \( a = -0.5 \, \text{m/s}^2 \) (deceleration), \( t = 3 \, \text{s} \).

Step2: Substitute the values

\( v = 5 + (-0.5)\times3 \)
\( v = 5 - 1.5 \)
\( v = 3.5 \, \text{m/s} \)

Answer:

\( \approx 2.45 \, \text{m/s} \)

Question 7