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dagg3r
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Hi guys I've got here motion questions and was wondering if someone could show me what i am doing wrong.
Q 1. A truck on a straight road starts from rest accelerating at 2.00m/s^2 until it reaches a speed of 20.0 m/s. then the truck travels for 20s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00s a) how long is the truck in motion
b) what is the average velocity for the motion described.
a) i added up the seconds ie 2m*10 + 20*20 + 5 = 35 seconds i got that right
b) this i got wrong i tell you what i did i broke up the av velocity into 3 sections so 1 section for the acceleration, 1 section for the 20s constant, and 5 seconds for the stop but got them wrong when i added then / 3 can someone just write down the steps and i will attempt to do this question thanks.
ie i did
step 1) average velocity = 10 m/s
v=u + at
v=0 + 2(10)
v=20 m/s
20 + 0 / 2 = 10m/s
step 2) average velocity= 20 m/s as velocity is constant
step 3) vf=0 vi=20
average velocity for section3 is 20/5=4 m/s
20+10+4 / 3 = 11.3 m/s that is incorrect the answer is 15.7 m/s
*** so someone please write down the steps and i will attempt to do it ok tahnks a lot every1 ***
Question 2 /
An inquisitive physics student and mountain climber climbs a 50m cliff that overhangs a calm pool of water. he throws 2 stones vertically downward, 1s apart, and observed that they cause a single splash. The first stone has an initital speed of 2 m/s a) How long after release of the first stone do the two stones hit the water b) what was the initital velocity of the second stone c) what is the velocity of each stone at the instant the two hit the water
ok for this question i could not do any of the parts i tried to find the time but was unsucessfull i will show you what i did.
i got confused with the time when the first stone dropped down so i use the formula x=ut + .5at^2
i put x as 2+ 2t thus 2 + 2 t = .5(-9.8)t^2
i solved for t but was incorrect can someone please tell me instructions / formulas to solve for parts a.b.c then i will try to attempt this and post back thank you very much
Q 1. A truck on a straight road starts from rest accelerating at 2.00m/s^2 until it reaches a speed of 20.0 m/s. then the truck travels for 20s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00s a) how long is the truck in motion
b) what is the average velocity for the motion described.
a) i added up the seconds ie 2m*10 + 20*20 + 5 = 35 seconds i got that right
b) this i got wrong i tell you what i did i broke up the av velocity into 3 sections so 1 section for the acceleration, 1 section for the 20s constant, and 5 seconds for the stop but got them wrong when i added then / 3 can someone just write down the steps and i will attempt to do this question thanks.
ie i did
step 1) average velocity = 10 m/s
v=u + at
v=0 + 2(10)
v=20 m/s
20 + 0 / 2 = 10m/s
step 2) average velocity= 20 m/s as velocity is constant
step 3) vf=0 vi=20
average velocity for section3 is 20/5=4 m/s
20+10+4 / 3 = 11.3 m/s that is incorrect the answer is 15.7 m/s
*** so someone please write down the steps and i will attempt to do it ok tahnks a lot every1 ***
Question 2 /
An inquisitive physics student and mountain climber climbs a 50m cliff that overhangs a calm pool of water. he throws 2 stones vertically downward, 1s apart, and observed that they cause a single splash. The first stone has an initital speed of 2 m/s a) How long after release of the first stone do the two stones hit the water b) what was the initital velocity of the second stone c) what is the velocity of each stone at the instant the two hit the water
ok for this question i could not do any of the parts i tried to find the time but was unsucessfull i will show you what i did.
i got confused with the time when the first stone dropped down so i use the formula x=ut + .5at^2
i put x as 2+ 2t thus 2 + 2 t = .5(-9.8)t^2
i solved for t but was incorrect can someone please tell me instructions / formulas to solve for parts a.b.c then i will try to attempt this and post back thank you very much