Question

A rightward-moving dragster accelerates from 0 m/s to 75 m/s in 4.5 s. Determine the magnitude of the average acceleration of the dragster

in m/s/a
337.5 m/s/s
-16.67 m/s/s
28.5 m/s/s
-337.5 m/s/s
16.67 m/s/s
-28.5 m/s/s

Answer 1

The magnitude of the average acceleration of the dragster is 16.67
`(m)/(s^(2) )`

Step-by-step explanation:

Acceleration is a vector quantity that relates changes in velocity to the time it takes to occur.

The average acceleration of a mobile is calculated using the following expression:

`acceleration=(change of velocity)/(time)=(vfinal - vinitial)/(time)`

In this way you calculate the average change in speed in the desired time interval.

The unit of measurement in the International System of acceleration is the meter per second squared (m/s²)

In this case:

• vfinal=75 m/s
• vinitial= 0 m/s
• time= 4.5 s

Replacing:

`acceleration=(75 m/s - 0 m/s)/(4.5 s)`

Solving:

acceleration= 16.67
`(m)/(s^(2) )`

The magnitude of the average acceleration of the dragster is 16.67
`(m)/(s^(2) )`