Starting from rest at 10m to the left of the origin (the starting line of a race), a go-kart undergoes constant acceleration of 2.46 m/s2. Make a table calculating the karts position and velocity at 10 second intervals from t=0 to 60 seconds. I have created the table but I need help with the equations to use to find the position and the velocity?
v = a t = 2.46 t
x = -10 + (1.23) t^2
To find the position and velocity of the go-kart at different time intervals, you can use the following equations of motion:
Position equation:
1. s = ut + (1/2)at^2
Velocity equation:
2. v = u + at
Where:
- s is the position (distance)
- u is the initial velocity (in this case 0 m/s as the go-kart starts from rest)
- t is the time in seconds
- a is the constant acceleration (2.46 m/s^2)
Now let's calculate the position and velocity at 10-second intervals from t=0 to 60 seconds:
-----------------------------------------------
| Time (s) | Position (m) | Velocity (m/s) |
-----------------------------------------------
| 0 | 10 | 0 |
-----------------------------------------------
| 10 | | |
-----------------------------------------------
| 20 | | |
-----------------------------------------------
| 30 | | |
-----------------------------------------------
| 40 | | |
-----------------------------------------------
| 50 | | |
-----------------------------------------------
| 60 | | |
-----------------------------------------------
At t=0 seconds, the go-kart starts from rest 10 meters to the left of the origin, so the position is 10 meters and the velocity is 0 m/s.
Now let's calculate the position and velocity at t=10 seconds:
Using equation 1:
s = ut + (1/2)at^2
s = 0 * 10 + (1/2) * 2.46 * (10^2)
s = 0 + (1/2) * 2.46 * 100
s = 0 + 1.23 * 100
s = 123 meters
Using equation 2:
v = u + at
v = 0 + 2.46 * 10
v = 24.6 m/s
Now fill in the table accordingly. Repeat the same process for subsequent time intervals until t=60 seconds.