When will the sled attain this speed?

[Sneakatone]

a bobsled slides down an icy track making of 29 degrees with the horizontal. Assume that the motion is frictionless.
a) how far must the bobsled slide in order to attain a speed of 91 km/h if initially at rest?
I converted km/h to 25.27 m/s and I used d = (v^2)/(2gsin(theta)) where d= 67.1 m, which is correct.

b)when will the speed be attained?
I tried d/v -> 67.1/25.27 and had 2.65 s and 25/67.1
But the answer is wrong. So I am stuck on part b.

 

[PeterO]

a bobsled slides down an icy track making of 29 degrees with the horizontal. Assume that the motion is frictionless.
a) how far must the bobsled slide in order to attain a speed of 91 km/h if initially at rest?
I converted km/h to 25.27 m/s and I used d = (v^2)/(2gsin(theta)) where d= 67.1 m, which is correct.

b)when will the speed be attained?
I tried d/v -> 67.1/25.27 and had 2.65 s and 25/67.1
But the answer is wrong. So I am stuck on part b.

d/v presumably means you want to use distance / time to calculate here. The speed you should be using then is the average speed. You have used the final speed.

 

[Sneakatone]

how would I find average speed based on the data?

 

[PeterO]

how would I find average speed based on the data?

For motion under constant acceleration, it is fortunate that the average velocity is simply the average of the initial and final velocities.
Although the path a bobsled takes if far from straight, it is not too wrong to consider that path as straight, and so the direction of velocity is always positive (or negative - depending which way you defined positive) so the average speed is the average of the initial and final speeds.

 

[Sneakatone]

so would initial speed equal zero so the average is just 25.27 m/s /2=12.63 m/s ?

 

[PeterO]

so would initial speed equal zero so the average is just 25.27 m/s /2=12.63 m/s ?

You could try that and see if you get the right answer. Be bold and try it!

 

[Sneakatone]

yes! I did 67/12.63 and had around 5 seconds. I didnt try at first because I did not want to run out of chances on the online HW. Thanks a lot!