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aliina [53]
3 years ago
13

a runner covers the last straight stretch of the race in 4 s. durning that time, he speeds up from 5 m/s to 9m/s what is the run

ners acceleration in this part of the race?
Physics
1 answer:
Sergio039 [100]3 years ago
5 0

Answer:

The runner's acceleration was 1\ m/s^2

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

v_f=v_o+at

Solving for a:

\displaystyle a=\frac{v_f-v_o}{t}

The runner speeds up from vo=5 m/s to vf=9 m/s in t=4 seconds, thus:

\displaystyle a=\frac{9-5}{4}

\displaystyle a=\frac{4}{4}=1

The runner's acceleration was 1\ m/s^2

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Answer:

B according to the picture

Explanation:

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2 years ago
An object accelerates at 32 m/s² when a force of 71 N is applied to it. What is the object’s mass? show your work
amid [387]

Answer:

Its  answer is 2.21 kg.

Explanation:

F =m × a

71 = m × 32

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2 years ago
A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

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