All categories

3 years ago

A car traveling at 45 km/h starts to brake, and comes to a stop over a distance of 18 m. Calculate the acceleration

Physics

1 answer:

lukranit [14]3 years ago

8 0

Answer:

Acceleration, $a=8.68\ m/s^2$

Explanation:

Given that,

Initial speed of a car, u = 45 km/h = 12.5 m/s

Final speed, v = 0 (as they comes to rest)

Distance, d = 18 m

We need to find the acceleration of the breaking car. Using third equation of motion as follows :

$v^2-u^2=2ad\\\\\text{Where a is acceleration of the car}\\\\a=\dfrac{v^2-u^2}{d}\\\\a=\dfrac{(12.5)^2}{18}\\\\a=8.68\ m/s^2$

So, the acceleration of the braking car is $8.68\ m/s^2$ .

You might be interested in

Can someone tell me the answer and the reason

spin [16.1K]

Answer:

(a) the blocks all had different masses.

Explanation:

The surface is smooth, therefore coefficient of friction is tending to zero.

Forces for each blocks varied from 6N to 8N to 7N to 5N

The blocks were made of different materials and different materials are going to have varying weight for the same size of block.

8 0

3 years ago

An Indy car is pushed down on the track by the Bernoulli effect. Where does the air flowing around the moving car travel the fas

GalinKa [24]

D
Reason: cause im built different way

6 0

3 years ago

Bacterial cells are microscopic. What do you think has happened so that after only a few days we can see a colony of bacteria on

wariber [46]

The startling answer is:

The bacteria ate the agar and reproduced (made more bacteria). There are so many of them now that you can see the blob of them.

5 0

3 years ago

A motorcycle, which has an initial linear speed of 8.0 m/s, decelerates to a speed of 2.2 m/s in 4.1 s. Each wheel has a radius

SVETLANKA909090 [29]

Answer:

Explanation:

Given

Initial linear speed $v_1=8 m/s$

initial angular velocity $\omega _1=\frac{v_1}{r}=\frac{8}{0.6}=13.33 rad/s$

Speed after 4.1 s is $v_2=2.2 m/s$

$\omega _2=\frac{2.2}{0.6}=3.66 rad/s$

using $\omega _2=\omega _1+\alpha t$

where $\alpha$ is angular acceleration

$3.66=13.33+\alpha \cdot 4.1$

$\alpha =-2.37 rad/s^2$ i.e. clockwise

(b)angular displacement

$\theta =\omega _1t+\frac{\alpha t^2}{2}$

$\theta =13.33\times 4.1-\frac{2.37\cdot 4.1^2}{2}$

$\theta =54.66-19.75$

$\theta =34.91 rad$

3 0

4 years ago

Anybody please help me with these questions... I really need help...

zhannawk [14.2K]

Remember, Your electric field lines drawn from positive charge (protons) and goes to negative charge (electron)

So, Your Answers would be:
15) B) Positive

16) A) Negative

17) B) Positive

18) A) Negative

19) B) Positive

20) B) Positive

Hope this helps!

6 0

3 years ago