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2 years ago

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A dragster and driver together have mass 942.9 kg . The dragster, starting from rest, attains a speed of 26.3 m/s in 0.61 s. Fin

d the average acceleration of the dragster during this time interval. Answer in units of m/s 2 . lawson (al52364) – forces – szymczak – (41202) 4 032 (part 2 of 3) 10.0 points What is the size of the average force on the dragster during this time interval? Answer in units of N.

1 answer:

pshichka [43]2 years ago

5 0

Answer:

a) $a = 43.11m/s^2$

b) F = 40648.42N

Explanation:

We will use kinematics to calculate the average acceleration:

$a = \frac{Vf-Vo}{t}$

$a = \frac{26.3-0}{0.61}$

$a = 43.11m/s^2$

By dynamics:

$F = m*a$

F = 942.9*43.11

F = 40648.42N

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<em>Answer: Option d.</em>

Explanation:

<u>Accelerated Motion </u>

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Where $v_f, v_o, t$ are the final speed, initial speed, and time taken to change them, respectively

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$\displaystyle \frac{x_1}{x_2}=\frac{1}{2}$

being $x_1$ the distance from p to q and $x_2$ the distance from q to r

It means that

$x_2=2x_1$

From the equation for speed

$v_q=v_p+at_1\ \ \ ..........[1]$

$v_r=v_q+at_2\ \ \ ..........[2]$

Replacing [1] into [2]

$v_r=v_p+at_1+at_2$

$v_r=v_p+a(t_1+t_2)$

Solving for a

$\displaystyle a=\frac{v_r-v_p}{t_1+t_2}\ \ \ .........[3]$

We now write the equation for both distances .

$\displaystyle x_1=v_pt_1+\frac{at_1^2}{2}$

$\displaystyle x_2=v_qt_2+\frac{at_2^2}{2}$

Using [1] again

$\displaystyle x_2=(v_p+at_1)t_2+\frac{at_2^2}{2}$

Since

$x_2=2x_1$

We have

$\displaystyle (v_p+at_1)t_2+\frac{at_2^2}{2}=2\left (v_pt_1+\frac{at_1^2}{2}\right )$

Operating

$\displaystyle v_pt_2+at_1t_2+\frac{at_2^2}{2}=2v_pt_1+at_1^2$

Rearranging

$\displaystyle v_pt_2-2v_pt_1=at_1^2-at_1t_2-\frac{at_2^2}{2}$

Factoring both sides

$\displaystyle v_p(t_2-2t_1)=\frac{a}{2}\left (2t_1^2-2t_1t_2-t_2^2 \right )$

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Operating and simplifying, we get a second-degree equation

$\displaystyle t_2^2+t_1t_2-2t_1^2=0$

Factoring

$(t_2-t_1)(t_2+2t_1)=0$

The only positive and valid answer is

$t_2=t_1$

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$\displaystyle \frac{t_1}{t_2}=1$

The option d. is correct

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