Ask question

Ask question

All categories

2 years ago

13

A car starts from rest and travels for 5.0 s with a uniform acceleration of +1.5 m/s2 . The driver then applies the brakes, caus

ing a uniform acceleration of –2.0 m/s2. If the brakes are applied for 3.0 s, how fast is the car going at the end of the braking period?

1 answer:

Svetlanka [38]2 years ago

7 0

The final velocity of the car is +1.5 m/s.

Explanation:

We have to divide the problem into two parts.

In the first part, the car starts from rest and accelerates for 5.0 s. We can find the final velocity of the car after this first part using the suvat equation:

$v = u +at$

where

v is the final velocity

u = 0 is the initial velocity

$a=+1.5 m/s^2$ is the acceleration

t = 5.0 s is the time

Substituting,

$v=0+(1.5)(5.0)=7.5 m/s$

In the second part, the brakes are applied, so the car decelerates for 3.0 s, and the final velocity is given by

$v' = v + at$

where

v' is the final velocity

v = 7.5 m/s is the velocity at the beginning of this part

$a=-2.0 m/s^2$ is the deceleration

t = 3.0 s is the time

Substituting,

$v'=7.5+(-2.0)(3.0)=+1.5 m/s$

So, the final velocity is +1.5 m/s.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

You might be interested in

Find the acceleration of the system and the tension in the ropes for the system shown. The table mass is 30 kg and the hanging m

marusya05 [52]

The system's tension is 616 N and acceleration is 5.6 $m / s^{2}$

<u>Explanation:</u>

From newton’s second law of motion which state that net force acting on a body is product of mass of a body and acceleration of a body which is given as,

$F_{n e t}=m_{t o t} \times a$

Where,

$F_{n e t}$ is net force acting on body

$m_{\mathrm{tot}}$ is mass of body

a is acceleration of body

Given values

Table mass (m) = 30 kg

Hanging mass (m) = 40 kg

$a=\frac{F_{n e t}}{m_{\mathrm{tot}}}=\frac{m \times g}{m_{\mathrm{tot}}}$

Put the value for m = hanging mass = 40 kg and $g=9.8 \mathrm{m} / \mathrm{s}^{2}$ , we get

$a=\frac{40 \times 9.8}{30+40}=\frac{392}{70}=5.6 \mathrm{m} / \mathrm{s}^{2}$

The tension in the ropes, $T=(m \times g)+(m \times a)$

Here, m as hanging mass

T = tension, N or $k g m / s^{2}$

m = mass, kg

g = gravitational force, $9.8 \mathrm{m} / \mathrm{s}^{2}$

a = acceleration, $m / s^{2}$

$T = (40 \times 9.8)+(40 \times 5.6) = 392+224 = 616 N$

3 0

2 years ago

A circuit has a current of 1.2 A. If the voltage decreases to one-third of its original amount while the resistance remains

beks73 [17]

G 3.6A I think so try this answer

7 0

2 years ago

Light shines through a single slit whose width is 5.7 x 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a

lilavasa [31]

Answer:

$\lambda = 570\ nm$

Explanation:

Given,

Width of slit, W = 5.7 x 10⁻⁴ m

Distance between central bright fringe, L = 4 m

distance between central bright fringe and first dark fringe, y = 4 mm

Diffraction angle

$tan \theta = \dfrac{y}{L}$

$tan \theta = \dfrac{4}{4\times 10^3}$

$\theta = 0.0572$

Now.

$W sin \theta = m \lambda$

m = 1

$5.7 \times 10^{-4} \times sin (0.0572) = 1 \times \lambda$

$\lambda = 569.99 \times 10^{-9}\ m$

$\lambda = 570\ nm$

4 0

3 years ago

A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in

Troyanec [42]

Answer:

The value is $\epsilon = 3.84 *10^{-5} \ V$

Explanation:

From the question we are told that

The diameter of the ring is $d = 18 \ mm = 0.018 \ m$

The length of the solenoid is $l = 25 \ cm = 0.25 \ m$

The diameter of the solenoid is $D = 5.0 \ cm = 0.05 \ m$

The number of turns is N = 1500

The change in current in the solenoid is $\Delta I = 20 \ A$

The time taken is $\Delta t = 1 \ s$

Generally the radius of the ring is

$r = \frac{d}{2}$

=> $r = \frac{0.018 }{2}$

=> $r = 0.009 \ m$

Generally the area of the ring is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * 0.009^2$

=> $A = 2.545 *10^{-4}\ m^2$

Generally the induced emf is mathematically represented as

$\epsilon = A * \frac{dB}{dt}$

Here

$\frac{dB }{dt} = \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi *10^{-7} \ N/A^2$

So

$\frac{dB }{dt} = 4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}$

=> $\frac{dB }{dt} = 0.150816\ T/s$

So

$\epsilon = 0.150816 * 2.545 *10^{-4}$

=> $\epsilon = 3.84 *10^{-5} \ V$

3 0

2 years ago

Which way does a river flow??

spin [16.1K]

Downwards - from uphill towards the lowlands and eventually into the sea.

8 0

2 years ago