Physics grade 11 Problem?
Can you explain a bit please and thanks :D
1
A driver was detected exceeding the posted speed limit of 60 km per hour, how many KM per hour would she have been traveling over the limit if she had covered the distance of 10 KM in 5 minutes?
2. The navigator on a yacht fixed his position 12.00 noon and estimates that the yacht has a distance of 87.5km to run before entering harbor. If the yacht is making a steady 10km/hr what is your estimated time of arrival.
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1. If she did 10k in 5 mins, she was travelling at (12 x 10) = 120kph. She was exceeding the limit by 60kph.
There are (12 x 5) mins in 1 hr. So the multiplier is 12.
2. (87.5/10) = 8.75 (8 3/4) hrs.
If the yacht can only do 10 km in 1 hr., you divide the distance to go by 10. Result = hrs.
1. 10 km ÷ 5 min = 10 km ÷ 1/12 hr = 120 km/hr
She was actually going twice the speed limit.
2. 87.5 km ÷ 10 km/hr = 8.75 hr or 8 hr 45 min
His estimated time of arrive would be 8:45 PM.
For the common area you need to apply the kinematic equation for speed that's Vf=Vi+at the place Vf is the main suitable speed, Vi is the preliminary %., a is the acceleration, and t is the time. in this situation the acceleration is -g, on condition that gravity is continuously accelerating the ball downwards at 9.8 m/s^2. So we've Vf=Vi-gt Now basically plug on your t's and be sure Vfa=25-9.8*a million=15.2 m/s Vfb=25-9.8*2=5.4 m/s Vfc=25-9.8*3= -4.4 m/s On area c, the ball is on its technique lower back off so the %. is destructive. for 2, you make the main of the equivalent equation, this time we are fixing for t. while a project asks for a max top, you already know the main suitable vertical %. would be 0 pondering the fact that it is going to be moving downwards each 2nd after that. Vf=Vi-gt 0=Vi-gt gt=Vi t=Vi/g t=30/9.8= 3.06 s