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

Answer:

Their deceleration is - 19 $\frac{m}{s^{2} }$

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 $\frac{m}{s^{2} }$.

The mathematical expression for the acceleration is:

$a=\frac{final speed - initial speed}{final instant - initial instant}$

In this case:

• final speed= 0 $\frac{m}{s}$

• initial speed= 2.8 $\frac{m}{s}$

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

Replacing:

$a=\frac{0 \frac{m}{s} - 2.8\frac{m}{s} }{0.15 s - 0s}$

a= -18. 667 $\frac{m}{s^{2} }$ ≅ -19 $\frac{m}{s^{2} }$

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 $\frac{m}{s^{2} }$