Answer :
Answer:
The average force exerted on the car during this time is 5,943 N
Explanation:
Given;
mass of the car, m = 849 kg
initial velocity of the car, u = 0
time of motion of the car, t = 5.00 s
final velocity of the car, v = 35 m/s
The average force exerted on the car during this time is given by;
[tex]F = ma \\\\F = \frac{m(v-u)}{t}\\\\F = \frac{849(35-0)}{5}\\\\F = \frac{849(35)}{5}\\\\ F = 849*7\\\\F = 5,943 \ N[/tex]
Therefore, the average force exerted on the car during this time is 5,943 N
Answer:
5943N
Let's say (+x) = eastward
Average horizontal acceleration
ax = vx -v0x/5.00s
= 35.0m/s-0/5.00s
= +7.09m/s
From here we apply the second law of newton
During this period average horizontal force acting on car
Summation x = max = (849kg)(+7.09m/s²)
= 5943N
+5.943x10³N
= 5.94kN east ward.