Tracy (of mass 49 kg) and Tom (of mass 77 kg) are standing at rest in the center of the roller rink, facing each other, free to move. Tracy pushes off Tom with her hands and remains in contact with Tom’s hands, applying a constant force for 0.8 s. Tracy moves 0.4 m during this time. When she stops pushing off Tom, she moves at a constant speed. What is Tracy’s constant acceleration during her time of contact with Tom? Answer in units of m/s 2 .

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CarliReifsteck CarliReifsteck · Answer 1 · 2019-12-03T08:31:45+01:00

Answer:

Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².

Explanation:

Given that,

Mass of Tracy = 49 kg

Mass of Tom = 77 kg

Distance = 0.4 m

Time = 0.8 s

We need to calculate Tracy’s constant acceleration during her time of contact with Tom

Using equation of motion

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

Where, s = distance

a = acceleration

t = time

Put the value into the formula

[tex]0.4=0+\dfrac{1}{2}a\times(0.8)^2[/tex]

[tex]a=\dfrac{0.4\times2}{(0.8)^2}[/tex]

[tex]a=1.25\ m/s^2[/tex]

Hence, Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².

andromache andromache · Answer 2 · 2022-02-15T14:37:59+01:00

Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².

Here we applied the motion equation i.e. shown below:

[tex]S = ut + \frac{1}{2}at^2[/tex]

Here s = distance

a = acceleration

t = time

So,

[tex]0.4 = \frac{1}{2} a \times (0.8)^2\\\\a = \frac{0.4\times 2}{(0.8)^2}[/tex]

= 1.25 m/s²

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