Post

AP Physics 1 Exam Question Solution

Created by @nathanedwards
 at November 1st 2023, 12:12:18 am.
AP_Physics_1-Questions
#kinematics
#ap-physics-1
#uniform-acceleration

AP Physics 1 Exam Question

A car moves in a straight line with an initial velocity of 20 m/s. The car then accelerates uniformly at a rate of 4 m/s^2 for a distance of 150 meters. Calculate the final velocity of the car and the time it takes to travel this distance.

Answer

To find the final velocity and time it takes for the car to travel the given distance, we can use the following kinematic equation:

vf2=vi2+2aΔxv_f^2 = v_i^2 + 2a \Delta x

where:

  • vfv_f is the final velocity
  • viv_i is the initial velocity
  • aa is the acceleration
  • Δx\Delta x is the distance

Given: vi=20 m/sv_i = 20 \, \text{m/s} a=4 m/s2a = 4 \, \text{m/s}^2 Δx=150 m\Delta x = 150 \, \text{m}

To find the final velocity, substitute the given values into the equation:

vf2=(20 m/s)2+2(4 m/s2)(150 m)v_f^2 = (20 \, \text{m/s})^2 + 2(4 \, \text{m/s}^2)(150 \, \text{m})

Simplifying the equation:

vf2=400 m2/s2+1200 m2/s2v_f^2 = 400 \, \text{m}^2/\text{s}^2 + 1200 \, \text{m}^2/\text{s}^2
vf2=1600 m2/s2v_f^2 = 1600 \, \text{m}^2/\text{s}^2

Taking the square root of both sides:

vf=1600 m2/s2v_f = \sqrt{1600 \, \text{m}^2/\text{s}^2}
vf=40 m/sv_f = 40 \, \text{m/s}

Thus, the final velocity of the car is 40 m/s.

To find the time it takes to travel the given distance, we can use the formula:

Δx=vit+12at2\Delta x = v_i t + \frac{1}{2} a t^2

Substituting the known values:

150 m=(20 m/s)t+12(4 m/s2)t2150 \, \text{m} = (20 \, \text{m/s}) t + \frac{1}{2} (4 \, \text{m/s}^2) t^2

This equation is a quadratic, so let's rearrange it into standard form:

2 m/s2t2+20 m/st−150 m=02 \, \text{m/s}^2 t^2 + 20 \, \text{m/s} t - 150 \, \text{m} = 0

To solve this quadratic equation, we can use the quadratic formula:

t=−b±b2−4ac2a t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Where:

  • a=2 m/s2a = 2 \, \text{m/s}^2
  • b=20 m/sb = 20 \, \text{m/s}
  • c=−150 mc = -150 \, \text{m}

Substituting the values into the formula:

t=−20 m/s±(20 m/s)2−4(2 m/s2)(−150 m)2(2 m/s2) t = \frac{-20 \, \text{m/s} \pm \sqrt{(20 \, \text{m/s})^2 - 4(2 \, \text{m/s}^2)(-150 \, \text{m})}}{2(2 \, \text{m/s}^2)}

Simplifying the equation:

t=−20 m/s±400 m2/s2+1200 m2/s24 m/s2 t = \frac{-20 \, \text{m/s} \pm \sqrt{400 \, \text{m}^2/\text{s}^2 + 1200 \, \text{m}^2/\text{s}^2}}{4 \, \text{m/s}^2}
t=−20 m/s±1600 m2/s24 m/s2 t = \frac{-20 \, \text{m/s} \pm \sqrt{1600 \, \text{m}^2/\text{s}^2}}{4 \, \text{m/s}^2}
t=−20 m/s±40 m/s4 m/s2 t = \frac{-20 \, \text{m/s} \pm 40 \, \text{m/s}}{4 \, \text{m/s}^2}

We have two solutions: [ t_1 = \frac{-20 , \text{m/s} + 40 , \text{m/s}}{4 , \text{m/s}^2} = 5 , \text{s}] [ t_2 = \frac{-20 , \text{m/s} - 40 , \text{m/s}}{4 , \text{m/s}^2} = -15 , \text{s}]

Since time cannot be negative, we discard the negative solution. Therefore, the time it takes for the car to travel the given distance is 5 seconds.

Thus, the final velocity of the car is 40 m/s and it takes 5 seconds to travel the distance of 150 meters.

Chat

Save To Bookmark