Stuck on this question!

I don't understand why t=5.9 and for part c can't you verify qhen t=11.77?

image-d451cd55-c643-4c05-9f9a-35804b615094706541460305866708-compressed.jpg.jpeg

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Reply 1

mqb2766

It helps to upload what youve tried.
That time value for a) is correct. You must know how to find the maximum velocity (calculus)?
For c) where does the t=11.77 come from? Why do you think you cant verify?

(edited 1 year ago)

When you differentiate to get acceleration your e^(0.3t) has changed sign on the exponent?
Looks something similar on your last line for s, but Im really not sure with all the crossing outs etc.

(edited 1 year ago)

Reply 4

Vincent404

I calculated t which was how long the athlete ran for. Why is t half the value in the answer?

Reply 5

mqb2766

Original post by Vincent404

I calculated t which was how long the athlete ran for. Why is t half the value in the answer?

Without being funny can you post your working etc without all the crossing outs.
Your expression for s looks about right, so Im not sure why youre saying t is half the value.

(edited 1 year ago)

Reply 6

Vincent404

In the mark scheme they use t as 5.9 to find the maximum velocity.

Original post by Vincent404

In the mark scheme they use t as 5.9 to find the maximum velocity.

In #4

When you differentiate to get acceleration your e^(0.3t) has changed sign on the exponent?

This is in line 2 of your working. The last term in your acceleration expression is incorrect.