Question

Please refer to the pictures

Answer 1

A. When
`t=0`, the particle has initial velocity 0.5 m/s.

B. I assume the particle's displacement after 50 s is what's wanted. This is given by the definite integral

`\displaystyle \int_0^(50) v(t) \, dt`

which is equal to the signed area under the curve. We can find the area by splitting the region up at
`t=20`; then the total area is equal to the area of two trapezoids,

• 1/2 (20 s) (0.5 m/s + 2.0 m/s) = 25 m
• 1/2 (2.0 m/s) ((50 s - 20 s) + (40 s - 20 s)) = 50 m

Then the total displacement is 25 + 50 = 75 m.

C. The average acceleration over the first 20.0 s is

`(v(20.0\,\mathrm s) - v_0)/(20.0\,\rm s) = (2.0(\rm m)/(\rm s) - 0.5(\rm m)/(\rm s))/(20.0\,\rm s) = \boxed{0.075(\rm m)/(\mathrm s^2)}`

D. Instantaneous acceleration at any time is equal to the slope of the tangent line to the curve at that time.

Since
`v(40\,\mathrm s)=2.0(\rm m)/(\rm s)` and
`v(50\,\mathrm s)=0`, the slope at
`t=45.0\,\rm s` is

`(v(50\,\mathrm s) - v(40\,\mathrm s))/(50\,\mathrm s - 40\,\mathrm s) = (0 - 2.0(\rm m)/(\rm s))/(10\,\rm s) = \boxed{-0.20(\rm m)/(\mathrm s^2)}`

Note that this is the same as the average acceleration over the last 10 s of motion. In this case acceleration is constant since velocity is linear over this time interval.

E. We know from part (D) that the particle's acceleration is -0.2 m/s² for all
`40\,\mathrm s < t < 50\,\mathrm s`, which eliminates C and D.

The velocity is constant over the interval
`20\,\mathrm s < t < 40\,\mathrm s`, which means no acceleration over this interval, which eliminates A and leaves B as the correct plot.