Ask question

Ask question

All categories

3 years ago

9

A car travelling at 15m/s comes to a rest in a distance of 14m when the brakes are applied. Calculate the deceleration of the ca

r
Marking brainliest :)

2 answers:

vlada-n [284]3 years ago

6 0

Answer:

force of the breaks is 6650 N, direction opposite to direction of movement

Explanation:

malfutka [58]3 years ago

5 0

Answer :

deceleration of the car is $a=8.03\ m/s^2$

Explanation:

It is given that,

Initial speed of the car, u = 15 m/s

Final speed of the car, v = 0 (it comes to rest)

Distance covered by the car, d = 14 m

Let a is the acceleration of the car. We know that the rate of change of velocity of an object is called its acceleration. Mathematically, it can be calculated as :

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{-(15)^2}{2\times 14}$

$a=-8.03\ m/s^2$

So, the deceleration of the car is $a=8.03\ m/s^2$ . Hence, this is the required solution.

You might be interested in

A stone is thrown vertically downward with an initial speed of 12.0 m/s from the top of a building. The stone takes 1.54 s to re

mr_godi [17]

(a) The stone travels a vertical distance <em>y</em> of

<em>y</em> = (12.0 m/s) <em>t</em> + 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the acceleration due to gravity. Note that this equation assume the downward direction to be positive, and that <em>y</em> = 0 corresponds to the height from which the stone is thrown.

So if it reaches the ground in <em>t</em> = 1.54 s, then the height of the building <em>y</em> is

<em>y</em> = (12.0 m/s) (1.54 s) + 1/2 (9.80 m/s²) (1.54 s)² ≈ 30.1 m

(b) The stone's (downward) velocity <em>v</em> at time <em>t </em>is

<em>v</em> = 12.0 m/s + <em>g t</em>

so that after <em>t</em> = 1.54 s, its velocity is

<em>v</em> = 12.0 m/s + (9.80 m/s²) (1.54 s) ≈ 27.1 m/s

(and of course, speed is the magnitude of velocity)

8 0

2 years ago

-------------------------------------------------------------------------------

mixas84 [53]

Answer:

nice!!! but did you know that geico can save you 15% or more on car insurance!

Explanation:

8 0

2 years ago

Read 2 more answers

Stars are called blackbody radiators because they

scoundrel [369]

The answer is D) absorb most of the light that falls on them

A blackbody is an object that absorbs most (theoretically all) of the light that falls on it. Stars are called black bodies because they demonstrate something close to this property. Why do they appear white and not black? As black body radiators heat up, they slowly turn from black to grey to white, according to the human eye. Since stars are incredibly hot, they appear white to us.

8 0

3 years ago

Which statement correctly describes magnetic field lines?

skelet666 [1.2K]

The answer is A. the fields lines never cross, if you bring another magnet near it, the lines work just compress

4 0

3 years ago

Read 2 more answers

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

a.

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

7 0

3 years ago