Answer:
112 m/s², 79.1°
Explanation:
In the x direction, given:
x₀ = 0 m
x = 19,500 cos 32.0° m
v₀ = 1810 cos 20.0° m/s
t = 9.20 s
Find: a
x = x₀ + v₀ t + ½ at²
19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²
a = 21.01 m/s²
In the y direction, given:
y₀ = 0 m
y = 19,500 sin 32.0° m
v₀ = 1810 sin 20.0° m/s
t = 9.20 s
Find: a
y = y₀ + v₀ t + ½ at²
19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²
a = 109.6 m/s²
The magnitude of the acceleration is:
a² = ax² + ay²
a² = (21.01)² + (109.6)²
a = 112 m/s²
And the direction is:
θ = atan(ay / ax)
θ = atan(109.6 / 21.01)
θ = 79.1°
The engine must produce an acceleration of 112 m/s²
The rate of change of the velocity with respect to time is known as the acceleration of the object. Generally, the unit of acceleration is considered as meter/seconds².
Newton's three equations of motion are only applicable for the constant acceleration, the slope of the velocity time graph represents the acceleration of any object.
As given in the problem a missile is moving 1810 m/s at a 20.0-degree angle. it needs to hit a target 19,500 m away in a 32.0-degree direction in 9.20 s.
let us first consider the horizontal component in the x direction
The distance covered in the horizontal direction
x = 19,500 cos 32.0° m
the velocity in the horizontal direction
u = 1810 cos 20.0° m/s
t = 9.20 s
For finding acceleration in the horizontal direction by using the second equation of motion
x = ut + 1/2at²
19,500× cos 32.0° = 0 + (1810× cos 20.0°) ×9.20+ 0.5×a ×9.20²
a = 21.01 m/s²
Similarly in the vertical direction
The distance covered in the vertical direction
Y= 19,500 sin 32.0° m
the velocity in the vertical direction
u= 1810 sin 20.0° m/s
t = 9.20 s
For finding acceleration in the vertical direction by using the second equation of motion
y = ut +1/2at²
19,500 sin 32.0° = 0 + (1810 sin 20.0°)× 9.20 + 0.5×a× 9.20²
a = 109.6 m/s²
The resultant acceleration
a² = (21.01)² + (109.6)²
a = 112 m/s²
Thus, the engine must produce an acceleration of 112 m/s²
Learn more about acceleration from here
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