Question

19. The velocity (V) of a particle varies with position (x).
as shown in the graph. lts acceleration when
x = 4 m will be

(1) – 8 m/s2
(2) 8 m/s2
(3) 2 m/s2
(4) -2 m/s2
​

Answer 1

Answer:

-8 m/s

Explanation:

$v^2 = u^2+2as$

Rearranging this to find a gives:

$\frac{v^{2} -u^2}{2s} =a$

Deceleration is constant between 3m and 5 m (as the graph is a straight line downwards)

Substituting in values gives:

$\frac{2^2-6^2}{4} =-8$