Question

simon is riding a bike at 12 km/h away from his friend keesha. he throws a ball at 5 km/h back to keesha, who is standing still on a sidewalk.
how fast would keesha say the ball is traveling?
○ 5 km/h
○ 7 km/h
○ 12 km/h
○ 17 km/h

Explanation:

Step1: Analyze relative motion

Simon is moving away from Keesha at 12 km/h, and he throws the ball back towards Keesha at 5 km/h (relative to Simon). To find the speed relative to Keesha, we need to consider the relative velocities. Since Simon is moving away, the ball's velocity relative to Keesha is the sum of Simon's speed (away) and the ball's speed towards Keesha (relative to Simon)? Wait, no—wait, Simon is moving at 12 km/h away from Keesha. The ball is thrown back (towards Keesha) at 5 km/h relative to Simon. So from Keesha's frame, the ball's speed is Simon's speed (12 km/h away) minus the ball's speed towards Simon? Wait, no, let's think again. If Simon is moving at \( v_{Simon} = 12 \) km/h away from Keesha, and the ball is thrown towards Keesha at \( v_{ball/Simon} = 5 \) km/h (relative to Simon), then the velocity of the ball relative to Keesha (\( v_{ball/Keesha} \)) is \( v_{Simon} - v_{ball/Simon} \)? Wait, no, direction: Simon is moving in the positive direction (away from Keesha), so his velocity is +12 km/h. The ball is thrown towards Keesha, so its velocity relative to Simon is -5 km/h (since towards Keesha is opposite to Simon's motion). Then, the velocity of the ball relative to Keesha is \( v_{ball/Keesha} = v_{Simon} + v_{ball/Simon} \) (using relative velocity formula: \( \vec{v}_{ball/Keesha} = \vec{v}_{ball/Simon} + \vec{v}_{Simon/Keesha} \)). Wait, \( \vec{v}_{Simon/Keesha} \) is +12 km/h (Simon moving away from Keesha). \( \vec{v}_{ball/Simon} \) is -5 km/h (ball moving towards Keesha relative to Simon). So \( \vec{v}_{ball/Keesha} = -5 + 12 = 7 \) km/h? Wait, no, that can't be. Wait, maybe I got the direction wrong. Let's take Keesha as the reference frame (stationary). Simon is moving at 12 km/h in the direction away from Keesha (let's say along the positive x-axis). The ball is thrown by Simon towards Keesha, so relative to Simon, the ball's velocity is -5 km/h (since towards Keesha is negative x-direction). Then, the velocity of the ball relative to Keesha is \( v_{ball} = v_{Simon} + v_{ball/Simon} \) (because \( v_{ball/Keesha} = v_{ball/Simon} + v_{Simon/Keesha} \)). So \( v_{ball} = (-5) + 12 = 7 \) km/h? Wait, but that would mean the ball is moving towards Keesha at 7 km/h? Wait, no, if Simon is moving away at 12, and the ball is thrown back at 5, then from Keesha's perspective, the ball is moving towards her at \( 12 - 5 = 7 \) km/h? Wait, maybe the initial analysis was wrong. Let's think of it as: Simon is moving away, so his speed is 12 km/h relative to Keesha. The ball is thrown towards Keesha at 5 km/h relative to Simon. So the ball's speed relative to Keesha is 12 (Simon's speed away) minus 5 (ball's speed towards Simon) = 7 km/h? Wait, no, if Simon is moving away, the distance between Simon and Keesha is increasing at 12 km/h. The ball is moving towards Keesha at 5 km/h relative to Simon, so the rate at which the ball is approaching Keesha is 12 (the rate Simon is moving away) minus 5 (the rate the ball is moving towards Simon)? Wait, no, that would be if the ball was moving in the same direction as Simon. Wait, no, the ball is moving opposite to Simon's direction. So Simon is moving at 12 km/h in direction A, the ball is moving at 5 km/h in direction -A (towards Keesha). So relative to Keesha, the ball's speed is 12 (Simon's speed in A) plus 5 (ball's speed in -A)? Wait, no, speed is a scalar here (since we're talking about how fast it's moving towards Keesha). Wait, maybe the correct way is: from Keesha's point of view, Simon is moving away at 12 km/h, and the ball is being thrown back towards her at 5 km/h relative to Simon. So the ball's speed towards Keesha is 12 (the speed Simon is moving away) minus 5 (the speed the ball is moving towards Simon)? No, that doesn't make sense. Wait, let's use relative velocity formula. Let’s define:

- Let \( v_{S/K} = 12 \) km/h (velocity of Simon relative to Keesha, away from Keesha)
- Let \( v_{B/S} = -5 \) km/h (velocity of Ball relative to Simon, towards Keesha, so negative if \( v_{S/K} \) is positive)

Then, velocity of Ball relative to Keesha (\( v_{B/K} \)) is \( v_{B/S} + v_{S/K} \) (by relative velocity: \( \vec{v}_{B/K} = \vec{v}_{B/S} + \vec{v}_{S/K} \))

So \( v_{B/K} = -5 + 12 = 7 \) km/h? Wait, but that would mean the ball is moving at 7 km/h towards Keesha? Wait, no, if Simon is moving away at 12, and the ball is thrown towards Keesha at 5 relative to Simon, then the ball's speed towards Keesha is 12 - 5 = 7? Wait, maybe I messed up the sign. Let's take Keesha as the origin. Simon's position at time t is \( x_S = 12t \) (since he's moving at 12 km/h away). The ball is thrown at time t=0 (let's say) from Simon's position, with velocity relative to Simon of -5 km/h (towards Keesha). So the ball's position at time t is \( x_B = x_S(0) + v_{B/S} \cdot t + v_{S/K} \cdot t \)? Wait, no, at time t, Simon's position is \( x_S = 12t \) (if he starts at x=0). The ball is thrown from Simon's position at time t=0, with velocity relative to Simon of -5 km/h, so the ball's velocity relative to the ground (Keesha) is \( 12 - 5 = 7 \) km/h? Wait, no, velocity relative to ground is velocity relative to Simon plus velocity of Simon relative to ground. So if Simon is moving at 12 km/h relative to ground, and the ball is moving at -5 km/h relative to Simon, then relative to ground, the ball is moving at 12 - 5 = 7 km/h? Wait, but that would mean the ball is moving towards Keesha at 7 km/h? Wait, no, 7 km/h towards Keesha? But Simon is moving away at 12, so the ball is moving towards Keesha at 7 km/h? Wait, maybe the answer is 7 km/h? Wait, the options are 5, 7, 12, 17. So 7 is an option. Let's check again.

Wait, maybe the problem is simpler: Simon is moving away at 12 km/h, and he throws the ball back (towards Keesha) at 5 km/h. So from Keesha's perspective, the ball's speed is 12 (Simon's speed away) plus 5 (ball's speed towards her)? No, that would be 17, but that's not right. Wait, no—if Simon is moving away, the ball is thrown towards Keesha, so the relative speed is 12 + 5? Wait, no, that would be if the ball was thrown in the same direction as Simon. Wait, I think I made a mistake in direction. Let's think of it as:

- Simon's speed relative to Keesha: 12 km/h (away)
- Ball's speed relative to Simon: 5 km/h (towards Keesha)

So the ball's speed relative to Keesha is Simon's speed (away) minus the ball's speed (towards Simon) = 12 - 5 = 7 km/h. Yes, that makes sense. Because Simon is moving away at 12, and the ball is moving towards Keesha at 5 relative to Simon, so the net speed towards Keesha is 12 - 5 = 7 km/h.

Step2: Confirm the calculation

So \( 12 - 5 = 7 \) km/h.

Answer:

7 km/h (the option with 7 km/h)