Can someone please help with this ***KINEMATICS*** question?

Can someone please help with this KINEMATICS question? enter image source here

2 Answers
Feb 27, 2018

a = 1 m/s^2 and u = 5.5 m/s

Explanation:

With constant acceleration during a time interval Deltat, the body's instantaneous velocity equals its average velocity when time is halfway through the time interval.

The body travels 10 m during the 1 s interval between t = 4 s and t = 5 s. The average velocity during that interval is

V_"ave" = 10 m/1 s = 10 m/s

Therefore the body's instantaneous velocity at t = 4.5 s was 10 m/s.

Using that same logic, the body's instantaneous velocity at t = 6.5 s was 12 m/s. We can now calculate acceleration.

a = (DeltaV)/(Deltat) = (12 m/s - 10 m/s)/(6.5 s - 4.5 s) = 1 m/s^2

Now the kinematic formula v = u + a*t will allow us to calculate u. I will use the time t = 4.5, therefore v will be 10 m/s.

v = u + a*t

10 m/s = u + 1 m/s^2*4.5 s

u = 10 m/s - 1 m/s^2*4.5 s = 10 m/s - 4.5 m/s = 5.5 m/s

I hope this helps,
Steve

Feb 27, 2018

"The answers are:"

\qquad \qquad \qquad \qquad \quad \ \ a =1 \ \ m/s^2 \qquad \qquad \ "and" \qquad \qquadv = 11/2 \ \ m/s \quad.

Explanation:

"For a body moving with constant acceleration" \ a, "along a straight line, the equation of motion -- "
"the position function" \ s(t), "is given by:"

\qquad \qquad \qquad \qquad s(t) \ = \ 1/2 a t^2 + v_0 t + s_ 0, \qquad "where:" \qquad \qquad \qquad \qquad \qquad \qquad \quad \ (I)

\quad a = "acceleration," quad v_0 = "initial velocity" quad s_0 = "initial position."

"[I don't know if you are allowed to assume this (!!) It is a"
"standard result for this type of motion. ]"

"At any rate, back to the above equation."

"We can take the origin of the straight line to be a location of"
"our choice on the line. Taking the point O as the origin, will"
"prove to be very nice. Let's also start the clock when the body"
"is at point O."

"So we have:"

\quad "position at time" \ t = 0 \ \ "is:" \quad "point O"; \qquad rArr \quad \ s_0 = 0.

\quad "velocity at time" \ t = 0 \ \ "is:" \quad "velocity at point O"; \qquad rArr \quad v_0 = u.

"Putting these into our equation of motion in (I), we have:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ s(t) \ = \ 1/2 a t^2 + v_0 t + s_ 0.

\qquad \qquad :. \qquad \qquad \qquad \qquad \ s(t) \ = \ 1/2 a t^2 + u t + 0

\qquad \qquad :. \qquad \qquad \qquad \qquad \ s(t) \ = \ 1/2 a t^2 + u t. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ (II)

"Now we are given: between 4 s and 5 s, the body traveled 10m."

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad s(5) - s(4) \ = \ 10. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (III)

"Now we are also given: between 6 s and 7 s, body traveled 12m."

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad s(7) - s(6) \ = \ 12. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ (IV)

"Using the equation of motion from (II), eqn. (III) becomes:"

\qquad \qquad :. \qquad \ [ 1/2 a (5)^2 + u (5) ] - [ 1/2 a (4)^2 + u (4) ] \ = \ 10.

\qquad \qquad :. \qquad \qquad \quad [ 25/2 a + 5 u ] - [ 16/2 a + 4 u ] \ = \ 10.

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \quad 9/2 a + u \ = \ 10.

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \ \ 9 a +2 u \ = \ 20. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ (V)

"Similarly, using it with eqn. (IV), we have:"

\qquad \qquad :. \qquad \ [ 1/2 a (7)^2 + u (7) ] - [ 1/2 a (6)^2 + u (6) ] \ = \ 12.

\qquad \qquad :. \qquad \qquad \quad [ 49/2 a + 7 u ] - [ 36/2 a + 6 u ] \ = \ 12.

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \quad 13/2 a + u \ = \ 12.

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \ 13 a +2 u \ = \ 24. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ (VI)

"So now we see eqns. (V) and (VI) form a" \ \ 2 xx 2 \ "system of"
"linear equations for the desired quantities:" \quad a, u."

"So, let's solve these:"

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \ \ 9 a +2 u \ = \ 20.

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \quad \ \ 13 a +2 u \ = \ 24.

"(VI)" \ - \ (V):" \qquad \qquad :. \qquad \qquad \ 4 a \ = \ 4.

\qquad \qquad \qquad \qquad \qquad \qquad :. \qquad \qquad \qquad \qquad \quad a \ = \ 1. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ (VII)

"Substituting" \ a = 1 \ "into eqn. (V), we get:"

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \ \ 9 (1) +2 u \ \ = \ 20.

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \qquad 9 +2 u \ = \ 20.

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ \ \ 2 u \ = \ 11.

\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \ \ \qquad \qquad \quad \ u \ = \ 11/2. \qquad \qquad \qquad \qquad \qquad \qquad \qquad (VII)

"And so we have our desired results:"

\qquad \qquad \qquad \qquad \qquad a =1 \ \ m/s^2 \qquad \qquad \ "and" \qquad \qquadv = 11/2 \ \ m/s \quad.