Respuesta :
Answer
given,
Mass of Kara's car = 1300 Kg
moving with speed = 11 m/s
time taken to stop = 0.14 s
final velocity = 0 m/s
distance between Lisa ford and Kara's car = 30 m
a) change in momentum of Kara's car
Δ P = m Δ v
[tex]\Delta P = m (v_f-v_i)[/tex]
[tex]\Delta P = 1300 (0 - 11)[/tex]
Δ P = - 1.43 x 10⁴ kg.m/s
b) impulse is equal to change in momentum of the car
I = - 1.43 x 10⁴ kg.m/s
c) magnitude of force experienced by Kara
I = F x t
I is impulse acting on the car
t is time
- 1.43 x 10⁴= F x 0.14
F = -1.021 x 10⁵ N
negative sign represents the direction of force
Answer:
Explanation:
Given that,
Mass of kara car m =1300kg
Velocity at which kara car was moving Vi =11m/s
The car stopped after t= 0.14sec
Therefore the final velocity is Vf = 0m/s
a. What is the change in momentum?
Change is momentum can be determine using
∆p = ∆MV
∆p = M∆V
∆p = m(Vf-Vi)
∆p = 1300(0-11)
∆p = 1300×-11
∆p = —14,300 kgm/s
The change in momentum of kara's car is —14,300kgm/s.
b. Impulse felt be kara car?
Impulse can be determine by using Newton second law of motion
Ft = ∆p
Impulse is Ft
I = ∆p = -14,300kgm/s.
Then, the impulse felt by Kara's car is -14,300kgm/s
c. Magnitude of force experienced by Kara's car?
From the impulse formula,
Ft = ∆p
Therefore,
F = ∆p/t
F = -14,300/0.14
F = -102,142.9N.
F ≈ —102,143N
The force experienced by kara's is -102,143N.
We can also determined the deceleration of Kara's car
From Newton second law
F=ma
Then, a = F/m
a = -102,143/1300
a = -78.57m/s²
The negative sign show deceleration
We can also calculate the distance the car moved before coming to halt.
Using equation of motion
X =ut+½at²
X = 11×0.14 + ½(-78.57) × 0.14²
X = 1.54 —0.77
X = 0.77m
