A particle travels along the x axis with a constant acceleration of +2.7 m/s2. At a time 4.2s following its start, it is at x = -108 m. At a time 5.7 s later it has a velocity of +13.0 m/s. Find its position at this later time.
Please use kinematic equations to solve.
since a = 2.7 m/s^2, at t=4.2, v=11.34m/s
Then its acceleration drops to (13.0-11.34)/5 = 0.332 m/s^2
So, starting at t=4.2,
x = -108 + 0.332t + 0.166 t^2
At 5.7s after the first phase, then x = -100.7
To solve this problem using kinematic equations, we can use the following equations of motion:
1. v = u + at (Equation 1)
2. s = ut + (1/2)at^2 (Equation 2)
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
- s is the displacement or position
Since we are given the acceleration, we can use Equation 1 to find the initial velocity u at time t = 4.2s:
v = u + at
13.0 m/s = u + (2.7 m/s^2)(4.2s)
Rearranging Equation 1, we get:
u = 13.0 m/s - (2.7 m/s^2)(4.2s)
Calculate u:
u = 13.0 m/s - 11.34 m/s
u = 1.66 m/s
Now, we can use Equation 2 to find the displacement s at time t = 5.7s:
s = ut + (1/2)at^2
s = (1.66 m/s)(5.7s) + (1/2)(2.7 m/s^2)(5.7s)^2
Simplify and calculate s:
s = 9.462 m + 0.5(2.7 m/s^2)(32.49 s^2)
s = 9.462 m + 43.9825 m
s = 53.4445 m
Therefore, the position of the particle at t = 5.7s is 53.4445 m.