Question

particle has a constant acceleration of 5.8 m/s2. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.) (a) If its initial velocity is 2.1 m/s, at what time (in s after t = 0) is its displacement 4.0 m? (No Response) 0.867 s (b) What is its speed at that time (in m/s)? (No Response) 7.13 m/s

Answer 1

Answer:

a) t = 0.86 sec

b) v = 7.128 m/s

Explanation:

Given data:

Constant acceleration = 5.8 m/s^2

Initial velocity = 2.1 m/s

Displacement = 4.0 m

kinematic equation is given as

$X -X_o = v_o \times t + \frac{1}{2} at^2$

$X- X_o = 4.0 m$

v = 2.1 m/s

a = 5.8 m/s^2

plugging all value in the above relation

$4 = 2.1t + \frac{1}{2} 5.8t^2$

$2.9t^2 + 2.1t - 4 = 0$

solve for t

$t = \frac{-2 \pm \sqrt{2.1^2 -(4\times 2.9 \times (-4)}}{2\times 2.9}$

t = 0.86 sec

b)  kinematic equation relating to velocity is given as

$v = v_0 + at$

solving for velocity

$v = 2.1 + 5..8\times 0.867$

v = 7.128 m/s