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Bas_tet [7]
3 years ago
5

A turtle starts at a velocity of 2 m/s and accelerates at a rate of 3 m/s2. How long will it take the turtle to reach a final ve

locity of 11 m/s?
Physics
1 answer:
luda_lava [24]3 years ago
6 0
U= 2m/s
a= 3m/s^2
v= 11m/s
t=?


We know that a= v-u/t
t = v-u/a
t = 11-2/3
= 9/3
=3 seconds


Hope it helps!
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Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car
lapo4ka [179]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

A_a(t)  = 5m/s^2

To get the velocity, we integrate over time:

V_a(t) = (5m/s^2)*t + V_0

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V_a(t) = (5m/s^2)*t

To get the position equation we integrate again over time:

P_a(t) = 0.5*(5m/s^2)*t^2 + P_0

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

P_a(t) = 0.5*(5m/s^2)*t^2

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

V_b(t) =20m/s

To get the position equation, we can integrate:

P_b(t) = (20m/s)*t + P_0

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

P_b(t) = (20m/s)*t + 100m

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

P_a(t) = P_b(t)

We can solve this for t.

0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)}  }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}

We only care for the positive solution, which is:

t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

V_a(t) = (5m/s^2)*11.48s = 57.4 m/s

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Wow! Very simple. Use the equation for momentum which is p=mv. You know your momentum and velocity, so do 120=m(4), now solve for m, which is 30. The mass of the child is 30 kg
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