Let's be clear: The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.
Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.
After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.
It's 3.6 meters per second less than my speed was
at 4:19 PM last Tuesday.
Does that tell you anything ?
Why not ?
Explanation:
Below is an attachment containing the solution.
Net force on the car=F=4.8 x 10³ N
Explanation:
mass of car= 1.2 x 10³ Kg
initial velocity= Vi=0
Final velocity= Vf= 20 m/s
time = t= 5 s
Using kinematic equation,
Vf= Vi + at
20= 0 + a (5)
5 a=20
a= 20/5
a= 4 m/s²
Now force is given by F = ma
F= 1.2 x 10³ (4)
F=4.8 x 10³ N