Ask question

Ask question

All categories

2 years ago

7

If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?

1 answer:

Law Incorporation [45]2 years ago

5 0

a) 4.8 m/s² his average acceleration during the first 2.5 s

b) 15 m distance he cover during the first 2.5 seconds.

c) 31 m/s his speed as he finished the race.

<h3>(a) How to calculate the Average Acceleration ?</h3>

Average acceleration is calculated by using Newtons first law of motion

v = u + at

where

u is initial velocity

v is final velocity

a is constant acceleration

t is time

Given u = 0m/s , v = 12 m/s and t = 2.5 sec

Therefore average acceleration is given by

$a=\frac{v}{t}$

$a=\frac{12}{2.5}$

a = 4.8 m/sec²

b) Here we will use newtons second law of motion

$s=ut+\frac{1}{2}at^{2}$

On substituting value we get

s = 15m

c) Here we will use newtons third law of motion

v² = u² + 2as

Here u = 0 , a = 4.8 m/s² and s = 100 m

Therefore

v² = 960

v = 31 m/s

Disclaimer: the question was given incomplete in the portal. Here is the complete question.

Question: Usain Bolt's 100m sprint Runner set the world record for the100 meter sprint during the 2009World Championships in Berlin with a time of 9.58 s, reaching a top speed of 12 m/s in about 2.5 s.

a. What was his average acceleration during the first 2.5 s?

b. What distance did he cover during the first 2.5 seconds, assuming his acceleration was constant?

c. If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?

Learn more about Newtons law of motion here:

brainly.com/question/12525794

#SPJ4

You might be interested in

Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

6 0

3 years ago

A falling object accelerates from -10.0 m/s to -30.0 m/s. how much time does it take?

Zepler [3.9K]

Answer:

2.04 s

Explanation:

v = at + v₀

(-30.0 m/s) = (-9.8 m/s²) t + (-10.0 m/s)

t = 2.04 s

8 0

4 years ago

Read 2 more answers

How long does it take an automobile traveling 66.7 km/h to become even with a car that is traveling in another lane at 52.7 km/h

tresset_1 [31]

Answer:

The time taken is $t = 32.5 \ s$

Explanation:

From the question we are told that

The speed of first car is $v_1 = 66.7 \ km/h = 18.3 \ m/s$

The speed of second car is $v_2 = 52.7 \ km/h = 14.64 \ m/s$

The initial distance of separation is $d = 119 \ m$

The distance covered by first car is mathematically represented as

$d_t = d_i + d_f$

Here $d_i$ is the initial distance which is 0 m/s

and $d_f$ is the final distance covered which is evaluated as $d_f = v_1 * t$

So

$d_t = 0 \ m/s + (v_1 * t )$

$d_t = 0 \ m/s + (18.3 * t )$

The distance covered by second car is mathematically represented as

$d_t = d_i + d_f$

Here $d_i$ is the initial distance which is 119 m

and $d_f$ is the final distance covered which is evaluated as $d_f = v_2* t$

$d_t = 119 + 14.64 * t$

Given that the two car are now in the same position we have that

$119 + 14.64 * t = 0 + (18.3 * t )$

$t = 32.5 \ s$

6 0

3 years ago

A block is sent up a frictionless ramp along which an x axis extends upward. The figure below gives the kinetic energy of the bl

ss7ja [257]

Kinetic energy =1/2 mv^2

<span>m=2ke/v^2 </span>

<span>m=2(34)/3.6^2 </span>

<span>m=5.24 </span>

<span>force normal = mg </span>
<span>=5.24 x 9.8 </span>
<span>force normal = 51.4N

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

</span>

5 0

3 years ago

The depth of a pond is 1.5m. Calculate the pressure caused by the water at the bottom of the pond ??

zepelin [54]

Answer:

Area=1.5(1.5)=2.25m^2

Force of gravity=10N

\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}

⟼Pressure=

Area

Force

\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}

⟼Pressure=

2.25

10

\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}

⟼Pressure=4.4Pa

5 0

3 years ago