Asked by Cryztal Celeste on Sep 23, 2024
Verified
A free-falling object is described by h=35t+12at2h = 35 t + \frac { 1 } { 2 } a t ^ { 2 }h=35t+21at2 , where h is the height above the ground, t is the amount of time the object is falling, and a is the acceleration. Solve the equation for a .
A) a=2(h−35t) ta = \frac { 2 ( h - 35 t ) } { t }a=t2(h−35t)
B) a=2(h+35t) t2a = \frac { 2 ( h + 35 t ) } { t ^ { 2 } }a=t22(h+35t)
C) a=2(h−35t) t2a = \frac { 2 ( h - 35 t ) } { t ^ { 2 } }a=t22(h−35t)
D) a=h−35tt2a = \frac { h - 35 t } { t ^ { 2 } }a=t2h−35t
E) a=h−35t2t2a = \frac { h - 35 t } { 2 t ^ { 2 } }a=2t2h−35t
Free-Falling Object
An object that is falling under the sole influence of gravity, with no forces like air resistance acting upon it.
Acceleration
The rate of change of velocity of an object with respect to time.
- Understand and manipulate formulas to isolate a variable.
Verified Answer
QE
Qasem Elsharqawiabout 6 hours ago
Final Answer :
C
Explanation :
Firstly, taking the derivative of h with respect to t gives the velocity, v:
v=dhdt=35+atv = \frac{dh}{dt} = 35+atv=dtdh=35+at
Then, taking the derivative of v with respect to t gives the acceleration, a:
a=dvdt=ddt(35+at)=ddt(35)+ddt(at)=0+a=aa = \frac{dv}{dt} = \frac{d}{dt}(35+at) = \frac{d}{dt}(35) + \frac{d}{dt}(at) = 0 + a = aa=dtdv=dtd(35+at)=dtd(35)+dtd(at)=0+a=a
Therefore, the acceleration is just the coefficient of t in the equation for h:
a=2(h−35t)t2a = \frac{2(h-35t)}{t^2}a=t22(h−35t)
which is choice C.
v=dhdt=35+atv = \frac{dh}{dt} = 35+atv=dtdh=35+at
Then, taking the derivative of v with respect to t gives the acceleration, a:
a=dvdt=ddt(35+at)=ddt(35)+ddt(at)=0+a=aa = \frac{dv}{dt} = \frac{d}{dt}(35+at) = \frac{d}{dt}(35) + \frac{d}{dt}(at) = 0 + a = aa=dtdv=dtd(35+at)=dtd(35)+dtd(at)=0+a=a
Therefore, the acceleration is just the coefficient of t in the equation for h:
a=2(h−35t)t2a = \frac{2(h-35t)}{t^2}a=t22(h−35t)
which is choice C.
Learning Objectives
- Understand and manipulate formulas to isolate a variable.