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. #