Answer :
Answer:
a) 11 m/s
b) 0.0564 s
Explanation:
Given:
m = 2100 kg
vi = 22 ..... m/s before collision
vf = 0 ......after collision to stop
Δs = 0.62 distance traveled after collision .. crumpling of truck
Part a
[tex]V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s[/tex]
Part b
[tex]vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s[/tex]
(a) The required average speed of the truck during collision is 11 m/s.
(b) The required time interval for the collision is 0.058 s.
(c) The required magnitude of the average force exerted by the wall on the truck is [tex]7.96 \times 10^{5} \;\rm N[/tex].
(d) The required ratio of the force on the truck and the gravitational force is 38.67 : 1.
(e) The required approximations are:
- Neglect the horizontal component of the force of the road on the truck tires.
- Assume a nearly constant force exerted by the wall, so the speed changes at a nearly constant rate.
The section of analysis that deals with the motion of any object in one dimension are known as linear kinematics. The terms such as speed, velocity, and acceleration are the variables under kinematics.
Given data:
The mass of the truck is, m = 2100 kg.
The speed of the truck is, v = 22 m/s.
The distance crumpled by the truck is, d = 0.62 m.
(a)
Since the truck is going to stop finally (v' = 0) therefore the average speed is calculated as,
[tex]v_{av.}=\dfrac{v - v'}{2}\\\\ v_{av.}=\dfrac{22 - 0}{2}\\\\ v_{av.}=11 \;\rm m/s[/tex]
Thus, the required average speed of the truck during collision is 11 m/s.
(b)
Now, apply the first kinematic equation of motion as,
[tex]v' = v + at[/tex]
Here, a is the linear acceleration and t is the time interval for the collision.
Solving as,
[tex]0 = 22 + at\\\\ a = -22/t[/tex]
Now, apply the second kinematic equation as,
[tex]v'^{2}=u^{2}+2ad\\\\ 0^{2}=22^{2}+2 \times \dfrac{-22}{t} \times 0.62\\\\ \dfrac{27.28}{t}=484\\\\ t = 0.058 \;\rm s[/tex]
Thus, we can conclude that the required time interval for the collision is 0.058 s.
(c)
The expression for the magnitude of average force exerted by wall on truck is,
[tex]F_{av.} = \dfrac{mv}{t}[/tex]
Solving as,
[tex]F_{av.}=\dfrac{2100 \times 22}{0.058}\\\\ F_{av.}=7.96 \times 10^{5} \;\rm N[/tex]
Thus, the required magnitude of the average force exerted by the wall on truck is [tex]7.96 \times 10^{5} \;\rm N[/tex].
(d)
The ratio of force of the truck and the gravitation force on the truck is,
[tex]= \dfrac{F_{av}}{mg}[/tex]
Here, g is the gravitational acceleration.
Solving as,
[tex]=\dfrac{7.96 \times 10^{5} \;\rm N}{2100 \times 9.8}\\\\ =38.67[/tex]
Thus, the required ratio of the force on the truck and the gravitational force is 38.67 : 1.
(e)
The approximations that were necessary to make these approximations include the negligence of the horizontal component of a force and the assumption of a constant force exerted by the wall.
This is because:
- The horizontal force may act as an unbalanced force.
- And the constant force by wall results in the variation of the speed of the truck at the constant rate.
Learn more about the linear kinematics here:
https://brainly.com/question/24486060