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zvonat [6]
2 years ago
7

A person riding a bike has an initial velocity of 10 meters per second and accelerates at a constant rate of 2 meters per second

^2. After 5 seconds, the person would travel ___ meters.
Physics
1 answer:
Montano1993 [528]2 years ago
7 0

After 5 seconds, the person would have travel 75 m

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

  • a is the acceleration
  • v is the final velocity
  • u is the initial velocity
  • t is the time

<h3>How to determine the final velocity</h3>

We'll begin by obtaining the final velocity of the person. This can be obtained as follow:

  • Initial velocity (u) = 10 m/s
  • Acceleration (a) = 2 m/s²
  • Time (t) = 5 s
  • Final velocity (v) = ?

a = (v – u) / t

2 = (v – 10) / 5

Cross multiply

v – 10 = 2 × 5

v – 10 = 10

Collect like terms

v = 10 + 10

v = 20 m/s

<h3>How to determine the distance travelled</h3>

The distance travelled can be obtained as follow:

  • Initial velocity (u) = 10 m/s
  • Final velocity (v) = 20 m/s
  • Acceleration (a) = 2 m/s²
  • Distance (s) = ?

v² = u² + 2as

Collect like terms

2as = v² - u²

Divide both sides by 2a

s = (v² - u²) / 2a

s = (20² - 10²) / (2 × 2)

s = 75 m

Thus, the distance travelled by the person is 75 m

Learn more about velocity:

brainly.com/question/3411682

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

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A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 85.0 m/s2
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Maximum height attained by the model rocket is 2172.87 m

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  • Initial speed of the model rocket = u = 0
  • acceleration of the model rocket = a\ =\ 85.0 m/s^2
  • time during the acceleration = t = 2.30 s

We have to consider the whole motion into two parts

In first part the rocket is moving with an acceleration of a = 85.0 m/s^2 for the time t = 2.30 s before the fuel abruptly runs out.

Let s_1 be the height attained by the rocket during this time intervel,

s_1\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s_1\ =\ 0\ +\ 0.5\times 85\times 2.30^2\\\Rightarrow s_1\ =\ 224.825\ m

And Final velocity at that point be v

\therefore v\ =\ u\ +\ at\\\Rightarrow v\ =\ 0\ +\ 85.0\times 2.3\\\Rightarrow v\ =\ 195.5\ m/s.

Now, in second part, after reaching the altitude of 224.825 m the fuel abruptly runs out. Therefore rocket is moving upward under the effect of gravitational acceleration,

Let 's_2' be the altitude attained by the rocket to reach at the maximum point after the rocket's fuel runs out,

At that insitant,

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  • a = -g\ =\ -9.81\ m/s^2
  • Final velocity of the rocket at the maximum altitude = v_f\ =\ 0

From the kinematics,

v^2\ =\ u^2\ +\ 2as\\\Rightarrow 0\ =\ u^2\ -\ 2gs_2\\\Rightarrow s_2\ =\ \dfrac{u^2}{2g}\\\Rightarrow s_2\ =\ \dfrac{195.5^2}{2\times 9.81}\\\Rightarrow s_2\ =\ 1948.02\ m

Hence the maximum altitude attained by the rocket from the ground is

s\ =\ s_1\ +\ s_2\ =\ 224.85\ +\ 1948.02\ =\ 2172.87\ m

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