Ask question

Ask question

All categories

DanielleElmas [232]

2 years ago

15

A car sitting at a red light begins to accelerate at 2 m/s2 when the light turns green. It continues with this acceleration unti

l it reaches a speed of 20 m/s. It then travels at this speed for another few minutes. How far does the car travel in the first 35 s after the light changes to green

1 answer:

jek_recluse [69]2 years ago

7 0

Answer:

x_total = 600 m

Explanation:

This is an exercise and kinematics, let's find the time it takes to reach the velocity 20 m / s

v = v₀ + a t

as part of rest v₀ = 0

t = v / a

t = 20/2

t = 10 s

let's find the distance traveled in this time

x₁ = vo t + ½ a t2

x₁ = 0 + ½ 2 10²

x₁ = 100 m

The remaining time is

t₂ = 35 - t

t₂ = 35 - 10

t₂ = 25 s

as in this range it has a constant speed

v = x₂ / t₂

x₂ = v t₂

x₂ = 20 25

x₂ = 500 m

the total distance traveled is

x_total = x₁ + x₂

x_total = 100 + 500

x_total = 600 m

You might be interested in

A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc

katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

$t=\frac{5.60km}{4.85km/hr}=1.155hr$

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

$V_{ds}=V_{river}+V{boat}$

$V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr$

and now up stream:

$V_{us}=V_{boat}-V{river}$

$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

$t_{ds}=\frac{x}{v_{ds}}$

$t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr$

and the time up stream:

$t_{us}=\frac{x}{v_{us}}$

$t_{us}=\frac{2.80km}{2,80km/hr}=1hr$

so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$

$t=\frac{5.60km}{5.60km/hr}$

so

t= 1hr

4 0

3 years ago

Read 2 more answers

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

jenyasd209 [6]

The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.

2. This is immediate from (2.2.7).

3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =

3 0

3 years ago

a group of students is asking people whether they use plastic bags. what are the students doing? a. stating a problem b. analyzi

notsponge [240]

I believe the correct answer from the choices listed above is option C. A group of students is asking people whether they use plastic bags. By doing such, the students are <span>collecting data. They are collecting data of how many people uses plastic bags. Hope this answers the question.</span>

6 0

3 years ago

Read 2 more answers

A cyclist maintains a constant velocity of 6.1 m/s headed away from point A. At some initial time, the cyclist is 242 m from poi

KengaRu [80]

Answer:

553.1m

Explanation:

When an object moves at constant velocity we can express this movement like V=x/t, where V is the velocity, x is the displacement and t is the time spent on it.

In that way, the expression x=V.t give us the displacement from t=0s until t=51s, but we have to sum the initial distance from the point A.

x=242m +V.t = 242m + (6.1m/s x 51s) = 553.1m

7 0

3 years ago

. When two speakers play different sounds with a constant frequency, beats are heard with a frequency of 5 Hz. If the frequency

morpeh [17]

The beat frequency is the concept required to develop this process. The phenomenon is generated when you have two waves but their frequencies do not differ greatly. So the beat frequency will be the difference between those two waves. For this case we have the difference and one of the waves, therefore,

$F_b = |F_1-F_2|$

Our values are given as,

$F_b = 5Hz$

$F_1 = 1050Hz$

Using the formula of the frequency I will have,

$F_b = |F_1-F_2|$

Replacing we have,

$5 = |1050-F_2|$

$F_2 = 1050 \pm 5$

$F_2 = 1045,1055 Hz$

The possible values are two: 1045Hz and 1055Hz

3 0

3 years ago