Question

An object is moving along the positive x axis with a uniform acceleration of 4m/s2 at time t=0s x=5m and v=3m/s.
a) what will be the velocity and the positions of the object at t=2s ?
b) what will be the position of the object when it has a velocity of 5m/s ?

Answer 1

(I) At t=0s,
v=u+at
v=3+(4×2)
v=11m/s
And, s=5+ut+(1/2)at^2
s=5÷(3×2)+8
s=19m
(II) v=u+at
v=5+(4×2)
v=13m/s
And, s=ut+(1/2)at^2
s=5+(5×2)+8
s=23m

Answer 2

Answer:

(a). The velocity and the positions of the object at t=2 s is 11 m/s and 19 m.

(b). The position of the object is 7 m.

Explanation:

Given that,

Acceleration $a = 4 m/s^2$

Velocity v = 3 m/s

Time t = 0 s

Position x = 5 m

(a).  The velocity and the positions of the object at t=2 s.

The velocity after 2 sec will be

Using equation of motion,

$v=u+at$

Put the value in the equation

$v=3+4\times2$

$v =11\ m/s$

Now, The distance traveled after 2 sec

Using third equation of motion

$v^2-u^2=2as$

$\dfrac{(v^2-u^2)}{2a}=s$

$s=\dfrac{11^2-3^2}{2\times4}$

$s= 14\ m$

So, the position will be

$x_{0}= x+s$

$x_{0}=5+14$

$x_{0}=19\ m$

(b). The position of the object when it has a velocity of 5 m/s.

Using third equation of motion,

$v^2-u^2=2as$

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

$s=\dfrac{5^2-3^2}{2\times4}$

$s= 2\ m$

So, the position will be

$x'=x+s$

$x'=5+2=7\ m$

Hence,

(a). The velocity and the positions of the object at t=2 s is 11 m/s and 19 m.

(b). The position of the object is 7 m.