Question

A particle moving with uniform acceleration along a straight line ABC crosses point A at t=0 with a velocity 12 m/s, B is 40 m away from A and Cis 64 m away from A. The particle passes B at t = 4 s.
a.After what time will the particle be at C?
b.What is its velocity at C?
c.When does the particle reach A again?
d.Locate the point where the particle reverses its direction of motion.
e.Find the distance covered by the particle in the first 15 s.

Answer 1

Let the acceleration of the particle be a.
For motion between A and B
u= 12 m/s, s = 40 m, t = 4 s
s = ut + (1/2)a$t^2$
$\Rightarrow$ 40 = 12 x 4 + (1/2) x a x $(4{)}^2$
a = -1 $m{s}^{-2}$
For motion between A and C
64= 12t + (1/2)(-l)$t^2$
=> $r^2$ - 24t +128 = 0
=>(t- 8) (t- 16) = 0
=> t = 8 s, 16 s
A.The particle will be at C twice, at t = 8 s and t = 16 s.
B.Velocity of the particle at C
At t = 8 s. velocity of the particle v = 12 + (-1) x 8
= 4 m/s.
As the acceleration of the particle is negative, it will retard as it moves along ABC. At a point beyond C (i.e., at D), the particle will come to rest momentarily, and then it will move backwards with increasing speed. Throughout the motion, the particle decelerates, its velocity decreases continuously, velocity varies as 12,
11,10,..., 2,1,0,-1,-2-10, -11, -12, etc. Only from A to D the motion is retarded, during which speed varies as 12,11,10,..., 2,1,0. Subsequently, the particle speeds up, speed changing as 1, 2, 3,..., etc. Notice the difference between deceleration and retardation.
When the particle reaches A again, its displacement = 0
$\therefore$ 0= 12 $\times$ t + (1/2)(-1)$t^2$
=> t = 0, 24 s
At t = 0, velocity at A = 12 m/s,
At t = 24 s, velocity at A is 12 + (-1) x 24 = -12 m/s.
D.The particle reverses the direction of motion at D. For the motion between A and D
u = 12 m/s, v = 0, a = -1 m$s^{-2}$. If AD = s, from the equation $v^2=u^2+2as$
$(0{)}^2=(12{)}^2$ + 2 x (-1) x s
s = 72 m
E. After 12 s the particle comes to rest momentarily at D after covering a distance of 72 m.
Distance in subsequent 3 s
= 0 x 3 + (1/2) x (-1) x $(3{)}^2$ = 4.5 m
$\therefore$ d = 72 m + 4.5 m = 76.5 m