Question

11. A car travels at 40 m/s, decelerates at a constant 4 m/s² until rest.

Determine speed and distance after decelerating in 10 seconds!

Answer 1

Speed is 0 m/s.

Distance is 200 meters.

Step-by-step explanation:

One-Dimensional Motion

In this scenario, we can treat speed as the absolute value of velocity and distance as absolute value of displacement since we are dealing with one-dimensional motion.

There’ll only exist one component of vector, by applying magnitude formula, it’ll return value as shown below:

Suppose we have a one-dimensional vector,
`\displaystyle{\vec v = 5\hat i}`, this represents a vector with magnitude of
`\displaystyle = √(5^2) = 5`.

Theorem I - Magnitude of One-Dimensional Vector

Suppose we have a vector
`\displaystyle{\vec v = a\hat i}` then the magnitude is
`\displaystyle{|\vec v| = \sqrt {a^2} = |a|}`.

Speed and Distance, Velocity and Displacement

Speed and distance both are scalar quantity which means they only have magnitude but lack the direction. In one-dimensional motion, speed is defined to be the absolute value of velocity and direction is defined to be the absolute value of displacement, the reason is shown above regarding one-dimensional vector.

An example would be, consider a car moves at velocity of -4 m/s, this means that a car just moves to negative direction or opposite direction of positive at 4 m/s. The negative and positive sign refer to the direction that an object moves, to know how fast they are moving, we consider its magnitude which is speed.

This can also be said same to displacement and distance, but be aware that this can only be applied to straight horizontal or vertical line movement or else the motion will become two-dimensional.

Solution

From the problem, there’s no specific distance or displacement’s graph so we will assume that the car is moving in straight horizontal line. The car is moving at speed of 40 m/s and decelerates at 4 m/s².

This means that our acceleration is -4 m/s² since it’s decelerating and is moving at 40 m/s from start. Therefore, we can apply a general physic formula which is:

`\displaystyle{v = u+at}`

Where v is final velocity, u is initial velocity, a is acceleration and t is time. As said, we can treat speed = positive velocity in this scenario so substitute u = 40 m/s and a = -4 m/s² in the formula.

`\displaystyle{v=40-4t}`

Now substitute t = 10 in the formula:

`\displaystyle{v = 40-4(10)}\\\\\displaystyle{v=40-40}\\\\\displaystyle{v=0 \ \sf{m/s}}`

Therefore, the speed of car at 10 seconds is 0 m/s, it’s not moving at this moment.

Next, we find distance which we can use the following formula:

`\displaystyle{s = ut + (1)/(2)at^2}`

Substitute u = 40, t = 10 and a = -4 in:

`\displaystyle{s=40(10)+(1)/(2)(-4)(10)^2}\\\\\displaystyle{s = 400-2(100)}\\\\\displaystyle{s=400-200}\\\\\displaystyle{s=200 \ \sf{m}}`

Therefore, in 10 seconds, it’ll be able to travel 200 meters.

Please let me know if you have any questions!