Question

The acceleration of a particle is defined by the relation a 5 28 m/s2. Knowing that x 5 20 m when t 5 4 s and that x 5 4 m when v 5 16 m/s, determine (a) the time when the velocity is zero, (b) the velocity and the total distance traveled when t 5 11 s.

Answer 1

Explanation:

It is given that,

The acceleration of a particle, $a=-8\ m/s^2$ (negative as the particle is decelerating)

Initial distance, x₁ = 20 m

Initial time, t₁ = 4 s

New distance x₂ = 4 m

Velocity, v = 10 m/s

(A) Calculating initial distance using second equation of motion as :

$x_1=ut_1+\dfrac{1}{2}at^2$

$20=4u+\dfrac{1}{2}(-8)\times 4^2$

u = 21 m/s

When velocity of the particle is zero, time taken is t (say). Using first equation of motion as :

$v=u+at$

$0=21+(-8)t$

t = 2.62 seconds

So, the velocity of the particle is zero at t = 2.62 seconds.

(B) Velocity at t = 11 s

$v=21+(-8)\times 11$

v = 13 m/s

Total distance covered at t = 11 s. The overall path travelled by the particle during its entire journey is called total distance covered.

$d=ut+\dfrac{1}{2}at^2+|ut+\dfrac{1}{2}at^2|$

$d=21\times 2.62+\dfrac{1}{2}\times (-8)(2.62)^2+|21\times 8.38+\dfrac{1}{2}\times (-8)(8.38)^2|$

d = 132.48 m

So, the distance travelled by the particle at t = 11 seconds is 132.48 meters.