Question

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

Answer 1

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:

`\boxed{\sf{\bold{a = (v_2 - v_1)/(t)}}}`

With the following condition :

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

Problem Solving

We know that :

• 
  `\sf{v_2}` = 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 :

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

Step by step :

`\sf{a = (v_2 - v_1)/(t)}`

`\sf{a * t = v_2 - v_1}`

`\sf{6.19 * 0.3 = 9.31 - v_1}`

`\sf{6.19 * 0.3 = 9.31 - v_1}`

`\sf{9.31 - v_1 = 1.857}`

`\sf{9.31 = 1.857 + v_1}`

`\sf{v_1 = 9.31 - 1.857}`

`\boxed{\sf{v_1 = 7.453 \: m/s}}`

Conclusion

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