Answer:
5.72 s
Explanation:
From Newton's law, F = ma
The East is +ve direction, Hence,
F = +8930 N
m = 2290 kg
a = ?
8930 = 2290 × a
a = 8930/2290 = 3.90 m/s²
So, we will find the time it takes the car to stop using the equations of motion
a = 3.90 m/s²
u = initial velocity of the car = - 22.3 m/s (the velocity is to the west)
v = final velocity of the car = 0 m/s (since the car comes to rest)
t = time taken for the car to come to rest = ?
v = u + at
0 = - 22.3 + (3.90)(t)
3.9t = 22.3
t = 5.72 s
I don’t know what book you’re talking about so I can’t help but have a look online, you may be able to find it if you search up the book name and look around a few websites
Answer:
the rate of acceleration of the train is 4 m/s²
Explanation:
Given;
initial velocity of the train, u = 10 m/s
change in time of motion, dt = 5 s
final velocity of the train, v = 30 m/s
The rate of acceleration of the train is calculated as;
Therefore, the rate of acceleration of the train is 4 m/s²
