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

Answer:

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

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

$v = u+at$

$a = \dfrac{v-u}{t}$

Where, v = final velocity

u = initial velocity

t = time

Put the value in the equation

$a = \dfrac{21.7-0}{5.05}$

$a = 4.297 m/s^2$

Now, using equation of motion again

For distance,

$s = ut+\dfrac{1}{2}at^2$

$s = 0+\dfrac{1}{2}\times4.3\times(5.05)^2$

$s=54.79\ m$

The distance is 54.79 m.

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

$v_{avg}=\dfrac{D}{T}$

Where, D = total distance

T = time

Put the value into the formula

$v_{avg}=\dfrac{54.79}{5.05}$

$v_{avg}=10.85\ m/s$

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