Ask question

Ask question

All categories

3 years ago

13

A racecar traveling at a velocity of 18.5 m/s, accelerates at a rate of 2.47 m/s2 and covers a distance of 79.78 m. Determine th

e final velocity of the racecar.

1 answer:

Vesna [10]3 years ago

4 0

Answer:

The final velocity of the race car is 27.14 m/s

Explanation:

Given;

initial velocity of the race car, u = 18.5 m/s

acceleration of the race car, a = 2.47 m/s²

distance covered by the race car, s = 79.78 m

Apply the following kinematic equation to determine the final velocity of the race car.

v² = u² + 2as

v² = (18.5)² + 2(2.47)(79.78)

v² = 736.363

v = √736.363

v = 27.14 m/s

Therefore, the final velocity of the racecar is 27.14 m/s

You might be interested in

A sound wave with a frequency of 400 Hz is moving through a solid object. If the wavelength of the sound wave is 8 m, what is th

Paraphin [41]

Answer:

v = 3200 m/s

Explanation:

As we know that the frequency of the sound wave is given as

$f = 400 Hz$

wavelength of the sound wave is given as

$\lambda = 8 m$

so now we have

$speed = wavelength \times frequency$

so we will have

$v = (8m) \times (400 Hz)$

$v = 3200 m/s$

4 0

2 years ago

A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the p

frez [133]

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

$a(t)=(5-10t)\ m/s^2$

(a) Since, $a=\dfrac{dv}{dt}$

v = velocity

$dv=a.dt$

$v=\int(a.dt)$

$v=\int(5-10t)(dt)$

$v=5t-\dfrac{10t^2}{2}=5t-5t^2$

(b) $v=\dfrac{dx}{dt}$

x = position

$x=\int v.dt$

$x=\int (5t-5t^2)dt$

$x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3$

(c) Velocity function is given by :

$v=5t-5t^2$

$5t-5t^2=0$

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

7 0

2 years ago

an object is dropped from a height of 25 meters. at what velocity will it hit the ground? a 7.0 m/s b 11 m/s c 22 m/s d 49 m/s e

kipiarov [429]

Assuming that the object starts at rest, we know the following values:

distance = 25m
acceleration = 9.81m/s^2 [down]
initial velocity = 0m/s

we want to find final velocity and we don't know the time it took, so we will use the kinematics equation without time in it:

Velocity final^2 = velocity initial^2 + 2 × acceleration × distance

Filling everythint in, we have:

Vf^2 = 0^2 + (2)(-9.81)(-25)
The reason why the values are negative is because they are going in the negative direction

Vf^2 = 490.5

Take the square root of that

Final velocity = 22.15m/s which is answer c

6 0

3 years ago

Remy wonders if the height of the mountain has anything to do with the eventual size of the tsunami wave. How should Remy test t

Anna71 [15]

Answer:

<em>The answer is B</em>

Explanation:

<em>I got this from study island</em>

5 0

3 years ago

Determine the acceleration that result when a 12-N net force is applied to a 3-kg object and then to a 6-kg object

user100 [1]

For 3 kg mass: <em>F = $\lpha ma,a$ = $\frac{F}{m}$ = $\frac{12N}{3kg}$ = 4 $\lpha m$ / $s^{2}$ </em>

For 6 kg mass: <em>F = $\lpha ma,a$ = $\frac{F}{m}$ = $\frac{12N}{6kg}$ = 2 $\lpha m$ / $s^{2}$ </em>

Hope this helped, have a great day!~

3 0

3 years ago