Physics, 07.04.2020 19:01 fqjenfiq6959

A motorcycle accelerates uniformly from rest and reaches a linear speed of 24.8 m/s in a time of 9.87 s. The radius of each tire is 0.287 m. What is the magnitude of the angular acceleration of each tire

Answers: 3

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Answers

Answer from: Kalle91106

8.756 rad/s²

Explanation:

Given that:

A motorcycle accelerates uniformly from rest, then initial velocity v_i = 0 m/s

It final velocity v_f = 24.8 m/s

time (t) = 9.87 s

radius (r) of each tire = 0.287 m

Firstly; the linear acceleration of the motor cycle is determined as follows:

a_T =(V_f - v_i)/t

=(24.8-0)/9.87

=2.513 m/s²

Then; the magnitude of angular acceleration

α = a_T /r

=2.513/0.287

=8.756 rad/s²

Answer from: natalie3734

8.75rad/s²

Explanation:

The tires of the motorcycle undergo a rolling motion. Therefore, the tangential acceleration, $a_{T}$ , of the tires is equal to their linear acceleration, a. i.e

$a_{T}$ = a --------------(i)

But, the tangential acceleration, $a_{T}$ , is the product of the angular acceleration, $\alpha$ , and the radius of the each of the tires. i.e

$a_{T}$ = r $\alpha$ ------------(ii)

Combine equations (i) and (ii) as follows;

a = r $\alpha$ --------------(iii)

Also, the linear acceleration, a, is given by;

a = $\frac{v - u}{t}$ ------------------(iv)

Where;

v = final linear speed of the tire

u = initial linear speed of the tire

t = time taken for the motion

Combine equations iii and iv as follows;

$\frac{v - u}{t}$ = r $\alpha$ ------------------(v)

From the question;

v = 24.8m/s

u = 0 (since the motorcycle accelerates from rest)

t = 9.87s

r = 0.287m

Substitute these values into equation (v) as follows;

$\frac{24.8 - 0}{9.87}$ = 0.287 $\alpha$

$\frac{24.8}{9.87}$ = 0.287 $\alpha$

2.51 = 0.287 $\alpha$

$\alpha$ = $\frac{2.51}{0.287}$

$\alpha$ = 8.75rad/s²

Therefore, the angular acceleration of each tire is 8.75rad/s²

Answer from: Quest

answer: i pick a

explanation:

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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A motorcycle accelerates uniformly from rest and reaches a linear speed of 24.8 m/s in a time of 9.8...

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