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Ad libitum [116K]
3 years ago
5

A driver of a car traveling 12 m/s applies the breaks. After 2 seconds the speed drops to 8 m/s what is the cars acceleration?

Physics
2 answers:
Afina-wow [57]3 years ago
6 0

Answer:

Deceleration of the car is 2\frac{m}{s^{2} }.

Explanation:

Given:

initial speed = 12 m/s

final speed = 8 m/s

time = 2 s

To find:

acceleration = ?

Formula used:

According to first kinematic equation:

v = u + a t

Where, v = final speed

u = initial speed

t = time

a = acceleration

Solution:

According to first kinematic equation:

v = u + a t

Where, v = final speed

u = initial speed

t = time

a = acceleration

8 = 12 + 2 a

-4 = 2 a

a = -2 \frac{m}{s^{2} }

The negative sign implies that it is deceleration. Thus, deceleration of the car is 2\frac{m}{s^{2} }.



Minchanka [31]3 years ago
5 0

Answer:

Acceleration value of car = -2 m/s^2

Explanation:

 We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Here a car is traveling at 12 m/s, so initial velocity = 12 m/s

After 2 seconds the speed drops to 8 m/s, so final velocity = 8 m/s

Time taken = 2 seconds

Substituting

        8 = 12 + a * 2

         2a = -4

         a = -2 m/s^2

 Acceleration value of car = -2 m/s^2

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3 years ago
What is the refractive index of air if light travels through it at 3.0 108 m/s?
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2 years ago
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
liq [111]

Answer:

A) total time = 55.5 seconds

B) average velocity = 25.27 m/s

Explanation:

It starts from rest, so initial velocity, u = 0 m/s

We are given;

acceleration; a = 2 m/s²

Final velocity; v = 31 m/s

From Newton's first law of motion,

v = u + at

So, 31 = 0 + 2t

t = 31/2

t = 15.5 sec

We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;

(a) Total time in which car is in motion = 15.5 + 35 +5 = 55.5 seconds

b)Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;

S = ut + ½at²

S1 = 0 + ½(2 * 15.5²)

S1 = 240.25 m

Distance traveled in 35 sec with with velocity of 31 m/sec is;

S2 = velocity x time

S2 = 35 × 31 = 1085 m

Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity(u) will be 31 m/s.

From the first equation of motion,

a = (v - u)/t

a = (0 - 31)/5

a = -6.2 m/s²

So, distance travelled is;

S3 = ut + ½at²

S3 = (31 × 5) + ½(-6.2 × 5²)

S3 = 155 - 77.5

S3 = 77.5 m

So overall total distance = S1 + S2 + S3

Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m

Average velocity = total distance/total time

Average velocity = 1402.75/55.5 = 25.27 m/s

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Answer:

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Explanation:

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