Question

Assume your car reaches a speed of 21.7 m/s at a steady rate for 5.05 s after the light turns green. (a) What distance have you traveled during this time? (b) What is your average speed during this time?

Answer 1

The distance and average speed are 54.79 m and 10.85 m.

Step-by-step explanation:

Given that,

Speed = 21.7 m/s

Time = 5.05 s

(a). We need to calculate the distance

Firstly we will find the acceleration

Using equation of motion

`v = u+at`

`a = (v-u)/(t)`

Where, v = final velocity

u = initial velocity

t = time

Put the value in the equation

`a = (21.7-0)/(5.05)`

`a = 4.297 m/s^2`

Now, using equation of motion again

For distance,

`s = ut+(1)/(2)at^2`

`s = 0+(1)/(2)*4.3*(5.05)^2`

`s=54.79\ m`

The distance is 54.79 m.

(b). We need to calculate the average speed during this time

`v_(avg)=(D)/(T)`

Where, D = total distance

T = time

Put the value into the formula

`v_(avg)=(54.79)/(5.05)`

`v_(avg)=10.85\ m/s`

Hence, The distance and average speed are 54.79 m and 10.85 m.