A dragster and driver together have mass 918.8 kg. The dragster,starting from rest,attains a speed of 25.8 m/s in .53s .
Find the average acceleration of the dragsters during this time interval. Answer in units of m/s^2.
What is the size of the average force on the dragster during this time interval?
Assume the driver has a mass of 73.6kg. What horizontal force does the seat exert on the driver?
Here is my work,but i don't know if i did it correctly:
Vf=Vi+a(delta t)
25.8=0+a(.53)
48.68=a
918.8(48.68)=44,726.5=F
73.6(48.68)=3,582.8N
a = 25.8 / 53 = 0.487m/s
F = 918.8*0.487 = 447.5N.
Fd = mg = 73.6kg * 9.8N/kg = 721.3N.
The angle between seat and hor = 0 deg.
Fh = 721.3sin(0) = 0 Newtons.
Fv = 721.3cos(0) = 721.3N = ver. force of driver.
To find the average acceleration of the dragster during the time interval, you correctly used the equation:
Vf = Vi + a(delta t)
Where:
- Vf is the final velocity (25.8 m/s),
- Vi is the initial velocity (0 m/s, since the dragster starts from rest),
- a is the average acceleration (which we want to find), and
- delta t is the time interval (0.53 s).
Solving for a, we have:
25.8 = 0 + a(0.53)
a = 25.8 / 0.53 ≈ 48.68 m/s²
So, the average acceleration of the dragster during this time interval is approximately 48.68 m/s².
Next, to find the average force on the dragster during this time interval, you correctly used the equation:
F = m*a
Where:
- F is the force (which we want to find),
- m is the mass of the dragster and driver (918.8 kg),
- a is the average acceleration (48.68 m/s²).
Plugging in the values, we have:
F = 918.8 * 48.68 ≈ 44,725.984 N
Therefore, the average force on the dragster during this time interval is approximately 44,726 N.
Finally, to find the horizontal force exerted by the seat on the driver, you can use a similar approach. Since the driver has a mass of 73.6 kg and experiences the same acceleration as the dragster, you can use the equation:
F = m*a
Where:
- F is the force exerted on the driver by the seat (which we want to find),
- m is the mass of the driver (73.6 kg),
- a is the average acceleration (48.68 m/s²).
Plugging in the values, we have:
F = 73.6 * 48.68 ≈ 3,582.848 N
Therefore, the horizontal force exerted by the seat on the driver is approximately 3,582.848 N.