Question

A particle moving with a constant acceleration has a velocity of 20 cm/s when its position is x = 10 cm. Its position 7.0 s later is x = −30 cm. What is the acceleration of the particle?

Answer 1

Given :

Initial velocity, v = 20cm/s.

Position, x = 10cm.

Time, t = 7 sec.

Later x becomes x = -30cm.

To find :

The acceleration of the particle.

Solution :

First, we will find the change in position.

$\implies \sf d_x \ = \ x_2 - x_1$

$\implies \sf -30 - 10$

$\implies \sf 40 cm$

Therefore, Change in position is 40cm.

Now,

Using the Newton's third equation to find the acceleration of the particle.

$\implies \sf d_x = v \times t + \dfrac {(at^2)}{2}$

$\implies \sf -40 = 20 \times 7 + \dfrac {49 a}{2}$

$\implies \sf -40 = 140 + \dfrac {49a}{2}$

$\implies \sf -80 = 280 + 49a$

$\implies \sf -49a = 28 + 80$

$\implies \sf -49a = 360$

$\implies \sf a \ = - \dfrac {360}{49} \ m/s^2$

$\implies \sf a \ \approx \ - 7.34 \ m/s^2$

Acceleration of the particle is -7.34 m/s².

Answer 2

$\Large\bf{\color{cyan}Given:}$

A particle moving with a constant acceleration has a velocity of 20cm/s when its position is x = 10cm.

$\Large\bf{\color{lime}To \ find:}$

The acceleration of the particle.

$\Large\bf{\color{peru}Solution:}$

Change in position of the particle = ?

$\large \leadsto \tt d_x = x_2 - x_1$

$\large \leadsto \tt -30-10$

$\large \leadsto \tt 40cm.$

Change in position is 40cm.

Hence,

Acceleration of the object = ?

By using the Newton's third equation :

$\large\color{indigo}{\underline{\boxed{\tt d_x = v \times t + \dfrac {at^2}{2}}}}$

$\large \leadsto \tt -40 = 20 \times 7 + \dfrac {49a}{2}$

$\large \leadsto \tt -40 = 140 + \dfrac {49a}{2}$

Multiplying by 2 on both sides,

$\large \leadsto \tt -80 = 280 + 49a$

$\large \leadsto \tt -49a = 28 + 80$

$\large \leadsto \tt -49 a = 360$

$\large \leadsto \tt a \ = - \dfrac {369}{49}$

$\large \leadsto {\red{\tt a \approx -7.34 \ m/s^2}}$

Therefore, acceleration of the particle is -7.34 m/s².