Are These the Correct Answers to this Kinematics Problem?

[physicsfun_12]

Homework Statement

Hello, I am currently revising for an exam later in the month, and was wondering if anyone could check my answers to this past examination question.

Thanks in advance.

The Displacement of a particle is given by x=20/3t³-23/2t²+6t+5 metres at time t. Determine:

a) the velocity and acceleration when t=0 seconds,
b) the velocity and acceleration when t=3 seconds,
c) the value of t when the body is at rest,
d) the value of t when the acceleration is 37 ms^-2
e) the distance traveled between t=2 seconds and t=3 seconds.

Homework Equations

d/dt(k.x^t)=k.tx^t-1

The Attempt at a Solution

I got answers as follows:

a) v=6 a=-23
b) v=105 a=97
c) 24.1 and 21.9
d) 1.5
e) 474

Any help with checking these answers will be greatly appreciated.

Thanks in advance

 

[Redbelly98]

(a) looks good.
(b) double-check v. a looks good.
. . . If you still get the same answer for v after trying again, please show your work.
(c) and (d) if you provide the expressions for v and a, and the equations you set up to solve, it will be easier for me or somebody else to check your final answers.
(e) I disagree. What values do you get for x at t=2 and t=3?

 

[physicsfun_12]

Hello, hope your well and thanks for your response.

For b) I got the same answer for v again. I just sub in t=3 to the velocity equation:

20*3² - 23*3 + 6 = 180 -75 = 105m/s.

For c) I said v=0 at rest so, 0 = 20t² - 23t + 6 = 0
and then used the quadractic formula to calculate t. Having done it again though, I got 28.02s and 17.98s this time

For d) I said a=37, so 37 = 40t -23 which solves to give t = 1.5.

For e) I got different answer again having done it again! I got x=2470.3m for t=2 and 99.5m for t=3. I then subtracted these to give a distance of 2370.8m

In summary, My new answers are as follows:

a) v=6, a =-23
b) v=105, a=97
c) 28.02 and 17.98
d) 1.5
e) 2370.8

Thanks again for your input

 

[ideasrule]

Hello, hope your well and thanks for your response.

For b) I got the same answer for v again. I just sub in t=3 to the velocity equation:

20*3² - 23*3 + 6 = 180 -75 = 105m/s.

-23*3+6 isn't -75; it's -63.

For c) I said v=0 at rest so, 0 = 20t² - 23t + 6 = 0
and then used the quadractic formula to calculate t. Having done it again though, I got 28.02s and 17.98s this time

Try it again. None of the answers you've given so far are right.

For d) I said a=37, so 37 = 40t -23 which solves to give t = 1.5.

That's right.

For e) I got different answer again having done it again! I got x=2470.3m for t=2 and 99.5m for t=3. I then subtracted these to give a distance of 2370.8m

99.5 m is right, but 2470.3 isn't. Try again. If you keep on making mistakes while punching numbers into your calculator, do it slowly or get a better calculator.

 

[Redbelly98]

For c) I said v=0 at rest so, 0 = 20t² - 23t + 6 = 0
and then used the quadractic formula to calculate t. Having done it again though, I got 28.02s and 17.98s this time

Look up the quadratic formula again, especially the part where you divide everything by 2a. Also, what did you get for the quantity

b² - 4ac ?​

 

[physicsfun_12]

Thanks ever so much for all your help. I Think I've got it now.

I was making a few silly errors.

For c) I wasn't dividing it all by 2a, just the bit in the square root!

For e) For some reason when I got down to the 53.33 - 46 + 12 +5 I multiplied 53.33 and 46 and then added the 12 and 5 to give that answer of 2470! Think I'd been working too long last night!

I think these answers are correct now (althougth wouldn't supprise me if they weren't lol!)

a) v=6m/s, a=-23m/s^2
b) 117m/s
c) 1.37s
d) 1.5s
e) 75.17m

Thanks again for eveyone's help