Respuesta :
Answer:
a) Acceleration of the car is given as
[tex]a_{car} = -21 m/s^2[/tex]
b) Acceleration of the truck is given as
[tex]a_{truck} = 10.15 m/s^2[/tex]
Explanation:
As we know that there is no external force in the direction of motion of truck and car
So here we can say that the momentum of the system before and after collision must be conserved
So here we will have
[tex]m_1v_1 + m_2v_2 = (m_1 + m_2)v[/tex]
now we have
[tex]1400 (6.32) + 2900(0) = (1400 + 2900) v[/tex]
[tex]v = 2.06 m/s[/tex]
a) For acceleration of car we know that it is rate of change in velocity of car
so we have
[tex]a_{car} = \frac{v_f - v_i}{t}[/tex]
[tex]a_{car} = \frac{2.06 - 6.32}{0.203}[/tex]
[tex]a_{car} = -21 m/s^2[/tex]
b) For acceleration of truck we will find the rate of change in velocity of the truck
so we have
[tex]a_{truck} = \frac{v_f - v_i}{t}[/tex]
[tex]a_{truck} = \frac{2.06 - 0}{0.203}[/tex]
[tex]a_{truck} = 10.15 m/s^2[/tex]
