the motion of a car along y axis is given by v(t)= -20 + 12 velocity v is in m/s and time in seconds find the instantaneous position of the car as a function of a time if at t=0 it was at 5 m. also find its acceleration at t=2 s
Gremory:
The first line v(t) = -20t :)
ya correct
thanks
^_^
:-)
Answers
Answered by
18
Hey
[tex]We \ know \ \frac{d}{dt}x = v \\ \\ So, for \ the \ instantaneous \ position \ of \ car :-\ \textgreater \ \\ \\ --\ \textgreater \ x(t) = \int\limits {(-20t + 12)} \; dt = -10t^2 + 12t + C \\ \\ Since, the object was at 5m at t = 0, C = 5[/tex]
Hence, Position of Object is given by -->
---> x(t) = -10t² + 12t + 5
Now,
[tex]Acceleration \ of \ body = \frac{d}{dt} v = \frac{d}{dt} ( -20t + 12 ) \\ \\ Hence, the \ particle's \ acceleration \ is \ a \ constant \ -20m s^{-2}[/tex]
Hope this helps you ^_^
[tex]We \ know \ \frac{d}{dt}x = v \\ \\ So, for \ the \ instantaneous \ position \ of \ car :-\ \textgreater \ \\ \\ --\ \textgreater \ x(t) = \int\limits {(-20t + 12)} \; dt = -10t^2 + 12t + C \\ \\ Since, the object was at 5m at t = 0, C = 5[/tex]
Hence, Position of Object is given by -->
---> x(t) = -10t² + 12t + 5
Now,
[tex]Acceleration \ of \ body = \frac{d}{dt} v = \frac{d}{dt} ( -20t + 12 ) \\ \\ Hence, the \ particle's \ acceleration \ is \ a \ constant \ -20m s^{-2}[/tex]
Hope this helps you ^_^
Similar questions