# The acceleration of gravity is 9.8 m/s

2
.
What is the magnitude of the total force on
a(n) 87 kg driver by the dragster he operates
as it accelerates horizontally along a straight
line from rest to 41 m/s in 5.3 s?

## Vo = 0

V = 41 m/s.
t = 5.3 s.
M = 87 kg

a = (V-Vo)/t

Fx = M*a

Fy = M*g

F = Sqrt(Fx^2+Fy^2) = Total force.

## To find the magnitude of the total force on the driver, we first need to determine the net force acting on the driver using Newton's second law of motion. According to this law, the net force (F_net) is equal to the product of mass (m) and acceleration (a).

Given:
Mass of the driver (m) = 87 kg
Initial velocity (u) = 0 m/s (rest)
Final velocity (v) = 41 m/s
Time (t) = 5.3 s

First, let's find the acceleration:
Acceleration (a) = (Change in velocity) / (Time)
a = (v - u) / t
a = (41 m/s - 0 m/s) / 5.3 s
a = 41 m/s / 5.3 s
a ≈ 7.74 m/s²

Now, let's calculate the net force:
F_net = m * a
F_net = 87 kg * 7.74 m/s²
F_net ≈ 673.38 N (Newtons)

Finally, to convert the net force from Newtons to kiloNewtons, we divide by 1000:
F_net = 673.38 N / 1000
F_net ≈ 0.67338 kN

Therefore, the magnitude of the total force on the 87 kg driver by the dragster is approximately 0.67338 kN.