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3 years ago

5

A motorcyclists starts from rest and reaches a speed of 6m/s after traveling with uniform acceleration for 3s .What is his accel

eration

1 answer:

siniylev [52]3 years ago

7 0

2m/s is the answer ! hopefully this helps

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A child swings on a playground swing. How many times does the child swing through the swing's equilibrium position during the co

Karolina [17]

<h2>The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>

Explanation:

One period means time taken to complete one revolution.

In case of swings in one period time it travels the same position through two times.

Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.

For 1 period = 2 times

For 3 periods = 3 x For 1 period

For 3 periods = 3 x 2 times

For 3 periods = 6 times

The child swing through the swing's equilibrium position 6 times during the course of 3 periods.

3 0

3 years ago

2- A student ran 135 meters in 15 seconds. What was the student's velocity?

RoseWind [281]

Answer:

9 Brainly hahaha ............huh

4 0

3 years ago

During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

geniusboy [140]

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

$a_t$ = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

$\omega_i$ = Initial angular velocity = 95 rad/s

$\theta$ = Angle of rotation

$\omega_f$ = Final angular velocity

t = Time taken

Angular acceleration is given by

$\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2$

The angular acceleration is -24.28571 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev$

The number of revolutions is 29.57239

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-24.28571}\\\Rightarrow t=3.91176\ s$

The time it takes for the car to stop is 3.91176 seconds

Linear distance

$s=r\theta\\\Rightarrow s=0.28\times 185.80885\\\Rightarrow s=52.026478\ m$

The distance the car travels is 52.026478 m

8 0

3 years ago

Proved that<br>V = u+at<br>

kondaur [170]

Answer:

$\sf Proof \ below$

Explanation:

We know that acceleration is change in velocity over time.

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

$\sf a=\frac{v-u}{t}$

v is the final velocity and u is the initial velocity.

Solve for v.

Multiply both sides by t.

$\sf at=v-u$

Add u to both sides.

$\sf at + u=v$

3 0

3 years ago

Read 2 more answers

A body travel according to the law x(t)=2t+3t3 calculate the accelaration of body at t=2s

faust18 [17]

Answer:

28m/s

Explanation:

Given law,

⇒ x(t) = 2t + 3t³

Where, t = 2s

Then,

⇒ x(2) = 2(2) + 3(2)³

⇒ 4 + 3(8)

⇒ 4 + 24

⇒ 28 m/s

Acceleration is 28m/s

4 0

3 years ago