Question

a body moving with uniform acceleration has a velocity of-11cm/s when its x coordinate is 3.00cm. if its x coordinate 2s later is -5cm, what is the magnitude in cm/s^2 of its acceleration ​

Answer 1

here it is confusing velocity of body is -11cm/s or 11cm/s. OK, I am solving for both cases.

case 1:- initial velocity of moving body, u = -11cm/s.

initial position of body , $s_0$ = 3cm

final position of body , s = -5cm

taken taken by body to move initial position to final position, t = 2s

using formula,

s = $\bf{s_0}$ + ut + 1/2 at²

or, -5cm = 3cm + (-11cm/s) × 2s + 1/2 × a(2s)²

or, -5 - 3 = -22 + 1/2 × a × 4

or, -8 + 22 = 2a

or, a = 7cm/s²

hence, acceleration of body will be 7cm/s²

if velocity is given -11cm/s , then magnitude of acceleration will be 7cm/s².

case2 : if initial velocity of body, u = 11 cm/s

initial position of body , $s_0$ = 3cm

final position of body , s = -5cm

taken taken by body to move initial position to final position, t = 2s

using formula,

s = $\bf{s_0}$ + ut + 1/2 at²

or, -5cm = 3cm + (11cm/s) × 2s + 1/2 × a(2s)²

or, -5 - 3 = 22 + 1/2 × a × 4

or, -8 - 22 = 2a

or, a = 15cm/s²

hence, if initial velocity of body is 11cm/s then, magnitude of acceleration will be 15 cm/s² .

Answer 2

Answer:

acceleration of moving body =a= -8 cm/s^2

Explanation:

velocity of moving body=v=11 cm/s

initial position of moving body=x1=3.00cm

final position of moving body=x2=-5cm

distance covered by  moving body=d=x2-x1

distance covered by moving body= -5-3=-8.cm

we know that

Distance=velocity/time

s=v/t

Distance=s= -8cm

velocity=v=11cm/s

time=t=?

time=velocity/distance covered

t=v/s

time=t=11/ -8= -11/8

acceleration of moving body = a =velocity/time=v/t

acceleration of moving body =a=-8 cm/s^2