What is the distance travel by the ball when the time is 1 sec?

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When a ball is thrown upwards, it reaches its highest point after 3 seconds, and the distance traveled in 1 second can be calculated using the formula for distance under constant acceleration. The initial speed was found to be 30 m/s, and the total distance over 3 seconds is 45 m, leading to an average speed of 15 m/s. However, the calculated distance for 1 second is 15 m, while the expected answer is 25 m, indicating a misunderstanding of the motion's dynamics. For the second question regarding a car decelerating from 20 m/s, the time to reach 100 m was calculated to be 10 seconds, confirming that average speed can be used for both scenarios due to constant acceleration. The discussion highlights the importance of correctly applying kinematic equations to solve motion problems.
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i throw a ball upwards
reach highest point at 3secs later
what is the distance travel by the ball when the time is 1 sec?
take g = 10m/s^2



another is
a car deccelerates from 20m/s at a rate of 2ms^2...
how long does it take to reach 100m

im trying to explain it to my younger brother. but I am not sure how to. can anyone help?
 
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huagulong said:
I'm trying to explain it to my younger brother. but I am not sure how to. can anyone help?
Really? Well, what has 'younger brother' tried to do already? Has 'he' looked at his class notes or text?
 
for the first question.

i manage to find the initial speed which was 30m/s. i derive it through the formulae

(v1-v2)/t = acc. so i drew out the v/t graph, and calcuate the area under graph when the time is 1sec. i got 25m.

the 2nd question. i use the same formulae (v1-v2)/t =acc. to find out the time. which is 10secs. if i calculate the area under graph, it's 100m. which is what the question is asking us to find.

is the method i used viable? or is there a simpler method for people to understand better?
 
i got another query. can i just find out the average speed for both of the question. and then use it to find the answer?

i can't get it for the 1st question. but i can get the answer for the 2nd one.
 
huagulong said:
i got another query. can i just find out the average speed for both of the question. and then use it to find the answer?
Yes, since the acceleration in both cases in constant, you can use the average velocity to compute the answer.
huagulong said:
i can't get it for the 1st question. but i can get the answer for the 2nd one.
If you post your attempts perhaps we can show you where your going wrong.
 
for the 1st question.:

i calculate out the initial speed which was 30m/s. and drew out the v/t graph.
using the graph i found out that the total distance traveled during the 3secs is 45m.
so the average speed would be 45/3 which is 15m/s.

i need to find out the distance traveled in 1sec. which would be 15m.
but the answer is 25m.
 
anyone?
 
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