The position x where the object comes to rest (momentarily) is
Acceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration (m / s²)v = final velocity (m / s)
u = initial velocity (m / s)
t = time taken (s)
d = distance (m)
Let us now tackle the problem!
This problem is about Kinematics.
Given:
vo = (-14.0i - 7.0j) m/s
a = (6.0i + 3.0j) m/s²
Unknown:
r = ? → v = 0 m/s
Solution:
To solve this problem, we need to use the following integral formula.
[tex]v = v_o + \int {a} \, dt[/tex]
[tex]v = (-14.0 \, \widehat{i} - 7.0 \, \widehat{j}) + \int {(6.0 \, \widehat{i} + 3.0\, \widehat{j})} \, dt[/tex]
[tex]v = (-14.0 + 6.0t) \, \widehat{i} + ( -7.0 + 3.0t ) \, \widehat{j}[/tex]
If the object comes to rest (momentarily) , then :
[tex]v_x = 0[/tex]
[tex](-14.0 + 6.0t) = 0[/tex]
[tex]6.0t = 14[/tex]
[tex]t = 14 \div 6.0[/tex]
[tex]\boxed {t = \frac {7}{3} ~ s}[/tex]
or
[tex]v_y = 0[/tex]
[tex]( -7.0 + 3.0t ) = 0[/tex]
[tex]3.0t = 7[/tex]
[tex]t = 7 \div 3.0[/tex]
[tex]\boxed {t = \frac {7}{3} ~ s}[/tex]
[tex]r = r_o + \int {v} \, dt[/tex]
[tex]r = 0 + \int { (-14.0 + 6.0t) \, \widehat{i} + ( -7.0 + 3.0t ) \, \widehat{j} } \, dt[/tex]
[tex]r = ( -14.0t + 3.0t^2 ) \, \widehat{i} + ( -7.0t + 1.5t^2 ) \, \widehat{j}[/tex]
At t = 7/3 s , then :
[tex]r = ( -14.0(\frac{7}{3}) + 3.0(\frac{7}{3})^2 ) \, \widehat{i} + ( -7.0(\frac{7}{3}) + 1.5(\frac{7}{3})^2 ) \, \widehat{j}[/tex]
[tex]r = [ (-16\frac{1}{3}) \, \widehat{i} + (-8\frac{1}{6}) \, \widehat{j} ] ~ m[/tex]
[tex]r \approx (-16.3 \, \widehat{i} - 8.2 \, \widehat{j} ) ~ m[/tex]
Grade: High School
Subject: Mathematics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Sperm , Whale , Travel
