A rocket powered sled accelerates a jet pilot in training straight forward from rest to 270 km/h in 12.1 seconds. Find:
a) The average acceleration of the sled
b) The time it takes to reach the speed limit on the highway, 100 km/h
c) The distance travelled when it reaches the final speed (270 km/h)

Answers

Answer 1

Answer:

6.198 m/s² 4.48 s 453.77 m

Explanation:

A Rocket Powered Sled Accelerates A Jet Pilot In Training Straight Forward From Rest To 270 Km/h In 12.1
Answer 2

a)The average acceleration of the sled will be 6.198 m/sec

b)The amount of time required to travel at the highway's 100 km/h speed limit will be 4.48 sec.

c)The distances covered when the vehicle hits its top speed of 270 km/h will be 453.77 m.

What is acceleration?

The rate of velocity change concerning time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.

[tex]\rm a = \frac{v-u}{t} \\\\[/tex]

The given data in the problem is given by ;

u is the initial speed of the rocket=  0 km/h

v is the final speed of the rocket=  270 km/h

t is the time interval= 12.1 seconds

a is the average acceleration of the ball=? m/sec²

to is the time it takes to reach the speed of 100 km/h

s is the distance traveled when it reaches the final speed

a)

The average acceleration of the sled is found as;

a = (v-u)/t

a=(75-0)/12.1

a = 6.198 m/sec

b)

The amount of time required to travel at the highway's 100 km/h speed limit;

From Newton's second equation of motion;

v =u+at

27.7 = 0 + 6.198t

t = 27.7 / 6.198

t = 4.48 sec

c)

The distances covered when the vehicle hits its top speed of 270 km/h;

v²=u²+2as

75²=0+2(6.198)s

s=(75)²/2(6.198)

s=453.77 m

Hence the average acceleration of the sled, the amount of time required to travel at the highway's 100 km/h speed limit, and the amounts of time required to travel at the highway's 100 km/h speed limit will be  6.198 m/sec,4.48 sec and 453.77 m respectively.

To learn more about acceleration, refer to the link;

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Answers

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

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We get;

[tex]v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]

[tex]v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]

The final relative velocity of the satellite, [tex]v_f[/tex] = v₁ + v₂

∴ [tex]v_f[/tex] = 0.095 + 0.095 = 0.190

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Answers

Answer:

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= 15-10/5

= 5/5

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Explanation:

hope this helped you.

Answer:

[tex]\boxed {\boxed {\sf 1 \ m/s^2}}[/tex]

Explanation:

Acceleration is the change in velocity over the change in time. Therefore, the formula for calculating acceleration is:

[tex]a= \frac{v_f-v_i}{t}[/tex]

Since the body's velocity increased from 10 meters per second to 15 meters per second 15 m/s is the final velocity and 10 m/s is the initial velocity. The time is 5 seconds.

[tex]v_f[/tex]= 15 m/s [tex]v_i[/tex]= 10 m/s t= 5 s

[tex]a= \frac{ 15 \ m/s - 10 \ m/s}{5 \ s}[/tex]

Solve the numerator.

15 m/s - 10 m/s = 5 m/s

[tex]a= \frac{ 5 \ m/s }{5 \ s}[/tex]

Divide.

[tex]a= 1 \ m/s/s[/tex]

[tex]a= 1 m/s^2[/tex]

The acceleration is 1 meter per second squared.

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