Question

Part A: A car is moving with a velocity of +21.0 m/s when a traffic light turns red. The driver hits the brakes and comes to a complete stop after traveling a distance of 32.5 m. What is the acceleration of the car during this event?Part B: How much time elapsed while the car was in the process of stopping?

Answer 1

Step-by-step explanation

Acceleration is the act of increasing or decreasing speed at a constant rate.

When an object is traveling in a straight line with an increase in velocity at equal intervals of time it is said to be in uniform accelerated motion

to find the acceleration and time we need to use the formulas:

`\begin{gathered} v_f=v_i+at \\ and \\ x=v_it+(1)/(2)at^2 \end{gathered}`

where

vf is the final velocity

vi is the initial velocity

t is the time

x is the traveled distance,

so

Step 1

Part A:

Given

`\begin{gathered} x=32.5\text{ m} \\ v_i=21(m)/(s) \\ v_f=0(rest) \end{gathered}`

so

a) replace the values in equation (1) and solve for a

`\begin{gathered} v_(f)=v_(i)+at \\ 0=21+at \\ -21=at \\ a=-(21)/(t) \end{gathered}`

b) now, replace in equation (2)

`\begin{gathered} x=v_(i)t+(1)/(2)at^(2) \\ 32.5=21t+(1)/(2)(-(21)/(t))t^2 \\ 32.5=21t-(21)/(2)t \\ 32.5=t(21-(21)/(2)) \\ 32.5=t(10.5) \\ t=(32.5)/(10.5)=3.09\text{ seconds} \end{gathered}`

som the time is 3.09 seconds

finally, replace in equation (1)

`\begin{gathered} v_(f)=v_(i)+at \\ 0=21+a(3.09) \\ -21=3.09a \\ a=-(21)/(3.09) \\ a=-6.78(m)/(s^2) \end{gathered}`

answers

accelerationi=-6.78 m/s^2

time= 3.09 seconds

I hope this helps you