A particle moves along X direction with constant acceleration. it velocity after 4s is 18 m/s and displacement is 56m find out its initial velocity and acceleration?

1 Answer
Jun 10, 2017

Initial velocity: 10"m"/"s"10ms

Acceleration: 2"m"/("s"^2)2ms2

Explanation:

We're asked to find the initial velocity v_(0x)v0x and the acceleration a_xax of an particle moving in one dimension with a known velocity and position at a certain time.

To find the initial velocity, we can use the equation

x = x_0 + ((v_x + v_(0x))/2)tx=x0+(vx+v0x2)t

We'll take the initial position x_0x0 to be 00. Let's rearrange this equation to solve for the initial velocity v_(0x)v0x:

x/t = (v_x + v_(0x))/2xt=vx+v0x2

v_(0x) = (2x)/t - v_xv0x=2xtvx

Plugging in known values, we have

v_(0x) = (2(56"m"))/(4"s") - 18"m"/"s" = color(red)(10"m"/"s"v0x=2(56m)4s18ms=10ms

The initial velocity is thus 10"m"/"s"10ms.

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Now that we know the initial velocity, we can use the equation

v_x = v_(0x) + a_xtvx=v0x+axt

to find its acceleration. Rearranging the equation to solve for a_xax, and plugging in known values, we have

a_x = (v_x - v_(0x))/t = (18"m"/"s" - 10"m"/"s")/(4"s") = color(blue)(2"m"/("s"^2)ax=vxv0xt=18ms10ms4s=2ms2

The acceleration of the particle is thus constant at 2"m"/("s"^2)2ms2.