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3 years ago

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A driver starts from rest on a straight test track that has markers every 0.14 km. The driver presses on the accelerator and for

the entire period of the test holds the car at constant acceleration. The car passes the 0.14 km post at 8.0 s after starting the test.
(a) What was the car's acceleration?
(b) What was the car's speed as it passed the 0.14 km post?

1 answer:

kari74 [83]3 years ago

5 0

Answer:

(A) Constant acceleration will be $a=1.479m/sec^2$

(B) Velocity of the car when it passes 0.14 km is 20.349 m/sec

Explanation:

It is given that driver starts from rest so initial velocity of the driver u = 0 m/sec

Distance traveled s = 0.14 km = 140 m

Time taken to cross 0.14 km is 8 sec

So time taken t = 8 sec

(A) From second equation of motion $s=ut+\frac{1}{2}at^2$

So $140=0\times 8+\frac{1}{2}\times a\times 8^2$

$a=1.479m/sec^2$

(B) From third equation of motion

$v^2=u^2+2as$

$v^2=0^2+2\times 1.479\times 140$

$v^2=414.12$

v = 20.349 m/sec

So speed of the car when it passed 0.14 km is 20.349 m/sec

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Answer:

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(B) is correct option.

Explanation:

Given that,

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Magnetic energy :

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(B) is correct option.

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A race car speeds up with an acceleration of 10 m/s2 for 21 s. If its final velocity is 210 m/s, find its initial velocity.

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The 100-m dash can be run by the best sprinters in 10.0 s. A 66-kg sprinter accelerates uniformly for the first 45 m to reach to

irga5000 [103]

(a) 154.5 N

Let's divide the motion of the sprinter in two parts:

- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration $a_1$ for a total time $t_1$ During this part of the motion, he covers a distance equal to $s_1 = 45 m$ , until he finally reaches a velocity of $v_1 = u + a_1t_1$ . We can use the following suvat equation:

$s_1 = u t_1 + \frac{1}{2}a_1t_1^2$

which reduces to

$s_1 = \frac{1}{2}a_1 t_1^2$ (1)

since u = 0.

- In the second part, he continues with constant speed $v_1 = a_1 t_1$ , covering a distance of $d_2 = 55 m$ in a time $t_2$ . This part of the motion is a uniform motion, so we can use the equation

$s_2 = v_1 t_2 = a_1 t_1 t_2$ (2)

We also know that the total time is 10.0 s, so

$t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)$

Therefore substituting into the 2nd equation

$s_2 = a_1 t_1 (10-t_1)$

From eq.(1) we find

$a_1 = \frac{2s_1}{t_1^2}$ (3)

And substituting into (2)

$s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1$

Solving for t,

$s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s$

So from (3) we find the acceleration in the first phase:

$a_1 = \frac{2(45)}{(6.2)^2}=2.34 m/s^2$

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The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation

$v_1 = u +a_1 t_1$

where we have

u = 0

$a_1 =2.34 m/s^2$ is the acceleration

$t_1 = 6.2 s$ is the time of the first part

Solving the equation,

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Suzie (of mass 42 kg) is roller-blading down the sidewalk going 32 miles per hour. She notices a group of workers down the walkw

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Answer:

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Substituting

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a = -6.46 m/s²

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Force = Mass x Acceleration

F = Ma

F = 42 x -6.46 =-271.52 N

Force exerted to stop Suzie = 271.52 N

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