Answer:
The velocity of the plane at take off is 160 m/s.
The distance travel by the plane in that time is 3200 meter.
Explanation:
Given:
Acceleration, a = 4 m/s²
Time, t = 40 s
u = 0 i .e initial velocity
To Find:
velocity , v = ?
distance , s =?
Solution:
we have first Kinematic equation
v = u + at
∴ v = 0 + 4×40
∴ v = 160 m/s
Now by Third Kinematic equation

∴ s = 0 + 0.5 × 4× 40²
∴ s = 3200 meter
Answer:
The change in kinetic energy (KE) of the car is more in the second case.
Explanation:
Let the mass of the car = m
initial velocity of the first case, u = 22 km/h = 6.11 m/s
final velocity of the first case, v = 32 km/h = 8.89 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(8.89² - 6.11²)
= 20.85m J
initial velocity of the second case, u = 32 km/h = 8.89 m/s
final velocity of the second case, v = 42 km/h = 11.67 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(11.67² - 8.89²)
= 28.58m J
The change in kinetic energy (KE) of the car is more in the second case.
Where is the rest .........