Solving a Particle's Motion with Acceleration a(t) = 3at + 2b

In summary, the given problem asks to determine the position and velocity of a particle with acceleration a(t)=3at+2b, given the values of a=1m/s3 and b=1m/s2. Using integration, the equations for velocity and position are found to be v(t)=3/2at2+2bt+1/2m/s and x(t)=1/2at3+bt2+1/2m/s(t)+1m, respectively. Substituting the given values for time t=1s, the velocity is found to be 4m/s and the position is 3m. The constant values for both equations are determined to be 1/2m/s and 1
  • #1
MelissaJL
50
0

Homework Statement


A particle moves with an acceleration a(t)=3at+2b with a=1m/s3 and b=1m/s2. Determine x(t) and v(t) such that v(1s)=4m/s and x(1s)=3m.

This is how I tried to solve it:
v(t)=∫(3at+2b)dt
v(t)=3/2at2+2bt+const
v(1s)=3/2m/s3(1s2)+2m/s2(1s)+const
4m/s=7/2m/s+const
const=1/2m/s

v(t)=3/2at2+2bt+1/2m/s

x(t)=∫(3/2at2+2bt+1/2m/s)dt
x(t)=1/2at3+bt2+1/2m/s(t)+const
x(1s)=1/2m/s3(1s3)+1m/s2(1s2)+1/2m/s(1s)+const
3m=1/2m+1m+1/2m+const
const=1m

x(t)=1/2at3+bt2+1/2m/s(t)+1m

Is this right, or where am I making errors?
 
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  • #2


Hello MelissaJL,

Welcome to Physics Forums!
MelissaJL said:

Homework Statement


A particle moves with an acceleration a(t)=3at+2b with a=1m/s3 and b=1m/s2. Determine x(t) and v(t) such that v(1s)=4m/s and x(1s)=3m.

This is how I tried to solve it:
v(t)=∫(3at+2b)dt
v(t)=3/2at2+2bt+const
v(1s)=3/2m/s3(1s2)+2m/s2(1s)+const
4m/s=7/2m/s+const
const=1/2m/s

v(t)=3/2at2+2bt+1/2m/s

x(t)=∫(3/2at2+2bt+1/2m/s)dt
x(t)=1/2at3+bt2+1/2m/s(t)+const
x(1s)=1/2m/s3(1s3)+1m/s2(1s2)+1/2m/s(1s)+const
3m=1/2m+1m+1/2m+const
const=1m

x(t)=1/2at3+bt2+1/2m/s(t)+1m

Is this right, or where am I making errors?
'Looks correct to me. :approve:
 

Related to Solving a Particle's Motion with Acceleration a(t) = 3at + 2b

1. What is the formula for acceleration?

The formula for acceleration is a = Δv/Δt, where a is acceleration, Δv is the change in velocity, and Δt is the change in time.

2. How do I solve a particle's motion with acceleration a(t) = 3at + 2b?

To solve a particle's motion with this specific acceleration function, you can use the equations of motion (v = v0 + at and x = x0 + v0t + 1/2at2) and plug in the given values for a(t), v0, and x0.

3. What are the units for acceleration?

The units for acceleration are typically meters per second squared (m/s2).

4. How does changing the values of a and b affect the particle's motion?

Changing the values of a and b in the acceleration function will directly affect the rate of change of the particle's velocity and position. A larger value for a will result in a faster change in velocity, while a larger value for b will result in a faster change in position.

5. Can this acceleration function be used to solve for non-uniform acceleration?

Yes, this acceleration function can be used to solve for non-uniform acceleration, as it includes both a linear term (3at) and a constant term (2b). However, it is important to note that this function assumes that the acceleration is constant with respect to time, and may not accurately model more complex acceleration patterns.

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