Question

A student is gliding along on a scooter at a comfortable 2.8 m/s when Mr. Jones walks around the corner and the two collide. If the student is brought to rest in 0.15 s, what is their acceleration (hint: should it be acceleration or deceleration)?

Answer 1

Their deceleration is - 19
`(m)/(s^(2) )`

Step-by-step explanation:

Acceleration is a quantity that indicates how the speed of the object changes over time. In other words, acceleration is a vector quantity that relates changes in velocity to the time it takes to occur. It is normally represented by the letter a and its unit of measurement, in the International System it is meters per second squared
`(m)/(s^(2) )`.

The mathematical expression for the acceleration is:

`a=(final speed - initial speed)/(final instant - initial instant)`

In this case:

• final speed= 0
  `(m)/(s)`

• initial speed= 2.8
  `(m)/(s)`

• final instant=0.15 s
• initial instant= 0 s

Replacing:

`a=(0 (m)/(s) - 2.8(m)/(s) )/(0.15 s - 0s)`

a= -18. 667
`(m)/(s^(2) )` ≅ -19
`(m)/(s^(2) )`

Deceleration is a quantity that expresses the passage of a moving body from one speed to a lower speed. Mathematically it is calculated as the acceleration but the result will be a negative acceleration.

Their deceleration is - 19
`(m)/(s^(2) )`