Respuesta :
Answer:
The car covers a distance of 68.93 meters.
Step-by-step explanation:
Given that,
Mass of the car, [tex]m=2.7\times 10^3\ kg[/tex]
Forward force acting on the car, [tex]F_1=1157\ N[/tex]
Force of friction acting on the car, [tex]f=902\ N[/tex]
The net force acting on the car is :
[tex]F=F_1-f\\\\F=1157-902\\\\F=255\ N[/tex]
Initial speed of the car, u = 0
Let d is the distance covered by the car so that its speed reaches to 3.6 m/s.
Force,
[tex]F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{255}{2.7\times 10^3}\\\\a=0.094\ m/s^2[/tex]
Now using third equation of motion as :
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2}{2a}\\\\d=\dfrac{(3.6)^2}{2\times 0.094}\\\\d=68.93\ m[/tex]
So, the car covers a distance of 68.93 meters. Hence, this is the required solution.
Answer:
Step-by-step explanation:
mass of car, m = 2.7 x 10^3 kg
Forward force, F = 1157 N
friction force, f = 902 N
initial velocity, u = 0 m/s
final velocity, v = 3.6 m/s
According to Newton's second law
F - f = m x a
where a is the acceleration of the car
1157 - 902 = 2.7 x 10^3 x a
a = 0.094 m/s²
Use third equation of motion
v² = u² + 2as
3.6 x 3.6 = 0 + 2 x 0.094 x s
s = 68.94 m
