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A particle has an acceleration of 6.19 m/s2 for 0.300 s. At the end of this time the particle's velocity is 9.31 m/s. Part A What was the particle's initial velocity

Sagot :

MDKPfromIDN

So, the particle's initial velocity was 7.453 m/s.

Introduction

Hi ! Here, I will help you to explain about the relationship between acceleration with changes in velocity and time. To be able to make changes in speed, a certain time interval is needed. If the detected change in velocity is negative, then the object will experience a deceleration. However, if the speed change detected is positive, then the object will experience an acceleration. Here is the equation that applies:

[tex] \boxed{\sf{\bold{a = \frac{v_2 - v_1}{t}}}} [/tex]

With the following condition :

  • a = acceleration (m/s²)
  • [tex] \sf{v_2} [/tex] = final speed of the object (m/s)
  • [tex] \sf{v_1} [/tex] = initial speed of the object (m/s)
  • t = interval of the time (s)

Problem Solving

We know that :

  • [tex] \sf{v_2} [/tex] = final speed of the object = 9.31 m/s
  • a = acceleration = 6.19 m/s²
  • t = interval of the time = 0.300 s

What was asked :

  • [tex] \sf{v_1} [/tex] = initial speed of the object = ... m/s

Step by step :

[tex] \sf{a = \frac{v_2 - v_1}{t}} [/tex]

[tex] \sf{a \times t = v_2 - v_1} [/tex]

[tex] \sf{6.19 \times 0.3 = 9.31 - v_1} [/tex]

[tex] \sf{6.19 \times 0.3 = 9.31 - v_1} [/tex]

[tex] \sf{9.31 - v_1 = 1.857} [/tex]

[tex] \sf{9.31 = 1.857 + v_1} [/tex]

[tex] \sf{v_1 = 9.31 - 1.857} [/tex]

[tex] \boxed{\sf{v_1 = 7.453 \: m/s}} [/tex]

Conclusion

So, the particle's initial velocity was 7.453 m/s.

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