I am having a problem answering a physics question: A plane is heading due south and climbing at the rate of 60 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 110 km/hr to the northwest, what is the ground speed of the plane?

To find the ground speed of the plane, we need to consider the effect of the wind on the plane's airspeed.

The airspeed of the plane is given as 540 km/hr. This means that in still air (no wind), the plane would be moving at this speed relative to the surrounding air.

However, there is a wind blowing at a speed of 110 km/hr to the northwest. This means that the wind is blowing in the opposite direction of the plane's heading.

To find the ground speed, we can use vector addition. We can break down the plane's airspeed and the wind speed into their respective north-south and east-west components.

The plane's airspeed has a north-south component of 0 km/hr (since it is heading due south) and an east-west component of 540 km/hr.

The wind speed has a north-south component of -110 km/hr (since it is blowing to the northwest, which is opposite to the direction of the plane's heading) and an east-west component of -110 km/hr.

To find the ground speed, we can add these components together.

The north-south component is 0 km/hr + (-110 km/hr) = -110 km/hr. The negative sign indicates a southward direction.

The east-west component is 540 km/hr + (-110 km/hr) = 430 km/hr.

To find the magnitude of the ground speed, we can use the Pythagorean theorem:

Ground speed = sqrt((-110 km/hr)^2 + (430 km/hr)^2 )

Ground speed = sqrt(12100 + 184900) = sqrt(197000) = 140.35 km/hr

Therefore, the ground speed of the plane is approximately 140.35 km/hr.

To find the ground speed of the plane, we need to consider both the airspeed of the plane and the effect of the wind.

The airspeed of the plane is given as 540 km/hr. This is the speed at which the plane would be moving if there were no wind.

The wind is blowing at a speed of 110 km/hr to the northwest. Since the plane is heading due south, we need to find the component of the wind's velocity that is acting against the plane's motion.

To do this, we can consider the velocity vectors involved. Let's break down the wind's velocity vector into its northward and westward components. Using simple trigonometry, we can determine that the northward component is -77.94 km/hr (opposite direction to the plane's motion) and the westward component is -77.94 km/hr (again, opposite direction to the plane's motion).

Since these components are acting against the plane's motion, we subtract them from the airspeed of the plane. So, the effective velocity of the plane with respect to the ground is:

Ground speed = Airspeed - Wind speed
= 540 km/hr - (-77.94 km/hr)
= 540 km/hr + 77.94 km/hr
= 617.94 km/hr

Therefore, the ground speed of the plane is approximately 617.94 km/hr.