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2 years ago

14

A bike accelerates uniformly from rest to a speed of 10 m/s over a distance of 50 m. (a) Determine the acceleration of the bike.

(b) how long will take to do that?

2 answers:

solong [7]2 years ago

8 0

Answer:

a) 0.2m/s

b) 5m/s

Explanation:

a) acceleration=∆v/∆t

v=10

t=50

10/50=1/5

=0.2m/s

b) time= d/s

d=50m

s=10

50/10

=5m/s

goldfiish [28.3K]2 years ago

8 0

Answer:

See below

Explanation:

Average speed = (0+ 10)/2 = 5 m/s

then to cover 50 m will take 50 m / 5 m/s = <u>10 seconds</u>

change in velocity/ change in time = <u>acceleration = 10/10 = 1 m/s^2 </u>

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$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

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$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

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a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

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