Explanation:
I assume the acceleration calculated in part (b) is the 3.33 m/s² from your other question.
Use Newton's second law to find the total force:
F = ma
F = (60,000 kg) (3.33 m/s²)
F = 200,000 N
Since there are 2 engines, the thrust from each is half of this:
F = 100,000 N
In reality, there are forces other than thrust. There are also drag forces (rolling friction and air resistance).
From Newton's second law, if we increase the mass and keep the force the same, the acceleration decreases. So it would take longer to reach the take-off speed.
Answer:
s = 27 m
Explanation:
given,
acceleration of the car, a = 6 m/s²
time of the travel, t = 3 s
final speed of the car, v = 0 m/s
using equation of motion
v = u + a t
0 = u - 6 x 3
u = 18 m/s
initial speed of the car = 18 m/s
Again using equation of motion
v² = u² + 2 a s
0 = 18² - 2 x 6 x s
12 s = 324
s = 27 m
Distance travel by the car is equal to 27 m.
Power = work /time
and W = force * dis
W = 96 * 3.046 J = 292.44 J
P = 292.44 / 4.6 = 63.575 watt