Knight riders physics thread

[knightrider]

Hi guys i am having trouble with these questions

calculate the instantaneous velocity of the motorcyclist at each
of the following times.
a) 15 s
b) 35 s

I know that i have to find the gradient at these times but when i try to i am getting confused

[knightrider]

bump

[Rishi97]

Hi guys i am having trouble with these questions

calculate the instantaneous velocity of the motorcyclist at each
of the following times.
a) 15 s
b) 35 s

I know that i have to find the gradient at these times but when i try to i am getting confused

For a, is the answer 20m/s ?

[Butterscotch]

Hi guys i am having trouble with these questions

calculate the instantaneous velocity of the motorcyclist at each
of the following times.
a) 15 s
b) 35 s

I know that i have to find the gradient at these times but when i try to i am getting confused

a) 20 ms^-1
b) -40 ms^-1

[Rishi97]

a) 20 ms^-1
b) -40 ms^-1

Thanks for confirming my answer.
Ok, so Knightrider, to find instantaneous velocity, as you know, we need to find the gradient. So using the rise/run formula, you should be able to find the gradient at these two points. You know, for the velocity at 15s, the gradient is increasing (positive) and at 35s, the gradient is decreasing(negative) so this is another way of checking your answers once you have found them.
Hope this helps

[knightrider]

Thanks guys

How would you do the following question

I got the distance to be 20km and the time is 2 hours. So i used speed=distance/time and got 20/2=10km/h

But the answers say 15km/h

[Zlatan]

Hey,

I'm assuming that the question is asking for the average speed for the whole journey. The formula that needed to be used was

average speed = total distance/ total time

You used 20 km as the total distance, but is that really the total distance covered ?? Notice how the question says she turns back and runs 5 km south after running 15 km north. At this point she has covered 20 km and is 10 km north from her starting point.

The question then says that after 2 hours she is 20 km from her starting point, implying that she has traveled 10 more kms north. Therefore after adding the values 20 km + 10 km, the total distance covered is 30 km, which when divided by 2 hours, gives you 15 km/h.

[knightrider]

Hey,

I'm assuming that the question is asking for the average speed for the whole journey. The formula that needed to be used was

average speed = total distance/ total time

You used 20 km as the total distance, but is that really the total distance covered ?? Notice how the question says she turns back and runs 5 km south after running 15 km north. At this point she has covered 20 km and is 10 km north from her starting point.

The question then says that after 2 hours she is 20 km from her starting point, implying that she has traveled 10 more kms north. Therefore after adding the values 20 km + 10 km, the total distance covered is 30 km, which when divided by 2 hours, gives you 15 km/h.

Thanks so for the starting point did you assume it is 0km

[Zlatan]

yeah, it would be safe to do that because that is where she "starts from" ..

 

[knightrider]

yeah, it would be safe to do that because that is where she "starts from" ..

 

Ok thanks so much 

[knightrider]

How would you do this question

What is the average velocity of the cyclist during this 11 s interval?

i have used velocity=dispacement/time and got -8/11 but the answers say i am wrong can someone help

[dankfrank420]

Is the answer 12 m/s?

I don't know how you got -8/11, but remember that the displacement is equal to the area under a velocity-time graph. Calculate the area under the graph (using squares/triangles) and you should get something like 132m.

v = d/t
= 132/11
= 12 m/s

[knightrider]

Is the answer 12 m/s?

I don't know how you got -8/11, but remember that the displacement is equal to the area under a velocity-time graph. Calculate the area under the graph (using squares/triangles) and you should get something like 132m.

v = d/t
= 132/11
= 12 m/s

Thankyou

[knightrider]

how would you do this question

Two physics students conduct the following experiment from a very high bridge.Thao drops a 1.5 kg shot-put from a vertical height of 60.0 m while at exactly the same time Benjamin throws a 100 g mass with an initial downwards velocity of 10.0 m s^−1 from a point 10.0 m above Thao.

At what time will the 100 g mass overtake the shot-put?

I think you have to solve equations simultaneously but what equations do you use and how do you get them

[Homer]

draw a velocity-time graph