Question

A car has an initial position of 5.8 m , an initial velocity of 2.9 m/s , and a constant acceleration of 0.73 m/s. What is the position of the car at the time of 2.9 s?

Answer 1

Given :

▪ Initial position of car = 5.8m

▪ Initial velocity of car = 2.9mps

▪ Acceleration of car = 0.73m/s^2

▪ Time interval = 2.9s

To Find :

▪ Position of car after given interval of time.

Solution :

✒ Since, acceleration of car has said to be constant throughout the motion, we can easily apply equation of kinematics to solve this type of questions.

✴ Second equation of kinematics :

☞ S = So + ut + (1/2)at^2

• S denotes distance (final position)
• So denotes initial position
• a denotes acceleration
• t denotes time
• u denotes initial velocity

____________________________

→ S = 5.8+(2.9×2.9)+0.5×0.73×(2.9)^2

→ S = 5.8+8.41+3.07

→ S = 17.28m

Answer 2

$\bigstar$ GIVEN :-

• Initial position of a car, $\sf{x_i}$ = 5.8 m

• Initial velocity of the car, u = 2.9 m/s

• Acceleration of the car. a = 0.73 m/s²

• Time, t = 2.9 s

$\bigstar$TO FIND :-

The position of the car at the time of 2.9 s.

$\bigstar$ACKNOWLEDGEMENT :-

As the acceleration is constant, so, the motion is uniform motion. For uniform motion, we know :

$\sf{\Delta x=ut+\dfrac{1}{2}at^2 }$

$\sf{\implies x_f-x_i=ut+\dfrac{1}{2}at^2}$

where, $\sf{x_f}$ is the final position and $\sf{x_i}$ is the initial position.

$\bigstar$ SOLUTION :-

Now, putting the known values in the above equation :-

$\sf{ x_f-x_i=ut+\dfrac{1}{2}at^2}\\\\\sf{\implies x_f-5.8=2.9\times 2.9+\dfrac{1}{2}\times 0.73(2.9)^2}\\\\\sf{\implies x_f=8.41+3.06 +5.8}\\\\\Large\boxed{\sf{\implies x_f=17.27\ m}}$

THEREFORE, THE POSITION OF THE CAR AT THE TIME 2.9 S IS 17.27 M.