Question

A pull back toy car accelerates for 1.8 seconds and covers 4 m while speeding up after being released from rest.

Answer 1

Acceleration, a = 2.47 seconds

Step-by-step explanation:

Given the following data;

Time, t = 1.8 seconds

Distance = 4 meters

Initial velocity = 0 m/s (since it's starting from rest)

To find the acceleration of the toy car, we would use the second equation of motion;

`S = ut + \frac {1}{2}at^(2)`

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

Substituting into the formula, we have;

`4 = 0(1.8) + \frac {1}{2}*a*(1.8)^(2)`

`4 = 0 + \frac {1}{2}*a*3.24`

`4 = \frac {1}{2}*a*3.24`

Cross multiplying, we have;

`8 = 3.24a`

`a = \frac {8}{3.24}`

Acceleration, a = 2.47 seconds