Variable acceleration in one dimension In a nutshell Sometimes an object will have variable acceleration. This means the acceleration of the object varies over time and can therefore be written as a function of time. These type of problems are similar to the ones which involve constant velocities or constant accelerations.
Equations DEscription Equation Displacement-velocity relation.
s = ∫ v d t v = d s d t \begin{aligned}s &= \int v \ dt \\\hspace{5mm}\\v&= \cfrac{ds}{dt}\end{aligned} s v = ∫ v d t = d t d s
Displacement-acceleration relation.
s = ∫ ( ∫ a d t ) d t a = d 2 s d t 2 \begin{aligned}s&= \int\left(\int a \ dt\right) dt\\\hspace{5mm}\\a &= \cfrac{d^2 s}{dt^2}\end{aligned} s a = ∫ ( ∫ a d t ) d t = d t 2 d 2 s
Velocity-acceleration relation.
v = ∫ a d t a = d v d t \begin{aligned}v &= \int a \ dt \\\hspace{5mm}\\a &= \cfrac{dv}{dt}\end{aligned} v a = ∫ a d t = d t d v
Variable definitions quantity name symbol unit name unit
D i s p l a c e m e n t Displacement D i s pl a ce m e n t
V e l o c i t y Velocity V e l oc i t y
M e t r e s p e r s e c o n d Metres \ per\ second M e t res p er seco n d
A c c e l e r a t i o n Acceleration A cce l er a t i o n
M e t r e s p e r s e c o n d s q u a r e d Metres \ per\ second \ squared M e t res p er seco n d s q u a re d
Variable acceleration in one dimension Example 1 A particle undergoes acceleration with a = ( 10 t + 5 2 ) m s − 2 a = \left(10t + \cfrac{5}{2}\right) \ ms^{-2} a = ( 10 t + 2 5 ) m s − 2 . It is given that v = 2 m s − 1 v = 2 \ m s^{-1} v = 2 m s − 1 t = 0 s . t=0 \ s. t = 0 s .
i) Calculate the velocity when t = 10 s . t=10 \ s. t = 10 s .
ii) Calculate the distance travelled after 10 s 10 \ s 10 s
i) Calculate the velocity when t = 10 s . t=10 \ s. t = 10 s .
Use v = ∫ a d t v = \int a \ dt v = ∫ a d t
v = ∫ a d t = ∫ ( 10 t + 5 2 ) d t = ∫ 10 t d t + ∫ 5 2 d t = 10 t 2 2 + 5 2 t + c = 5 t 2 + 5 2 t + c \begin{aligned}v &= \int a \ dt \\&= \int \left(10 t +\cfrac{5}{2}\right) \ dt \\&= \int 10 t \ dt + \int \cfrac{5}{2} \ dt \\& = \cfrac{10t^2}{2} + \dfrac52t + c \\&= 5t^2 + \dfrac52t + c\end{aligned} v = ∫ a d t = ∫ ( 10 t + 2 5 ) d t = ∫ 10 t d t + ∫ 2 5 d t = 2 10 t 2 + 2 5 t + c = 5 t 2 + 2 5 t + c
Given that v = 2 m s − 1 v=2\ ms^{-1} v = 2 m s − 1 t = 0 s t=0 \ s t = 0 s c . c. c .
2 = 5 × 0 2 + 5 2 × 0 + c c = 2 \begin{aligned}2 &= 5 \times 0^2 + \frac 52 \times 0 + c\\c &= 2\end{aligned} 2 c = 5 × 0 2 + 2 5 × 0 + c = 2
Therefore, the expression for the velocity as a function of time is:
v ( t ) = 5 t 2 + 5 2 t + 2 v(t) = 5t^2 + \frac 52 t + 2 v ( t ) = 5 t 2 + 2 5 t + 2
Substitute t = 10 s . t=10 \ s. t = 10 s .
v = 527 m s − 1 ‾ \underline{v = 527 \ ms^{-1}} v = 527 m s − 1
ii) Calculate the distance travelled after t = 10 s t=10 \ s t = 10 s
Use s = ∫ v d t s = \int v \ dt s = ∫ v d t
s = ∫ t 0 t v d t = ∫ t 0 = 0 s t = 10 s ( 5 t 2 + 5 2 t + 2 ) d t = [ 5 t 3 3 + 5 t 2 4 + 2 t ] 0 10 = 5000 3 + 500 4 + 20 = 1811.667.. m \begin{aligned}s &= \int_{t_0}^{t} v \ dt \\&= \int_{t_0=0\ s}^{t= 10 \ s} (5t^2 +\frac 5 2t + 2)\ dt \\&=\left[ \cfrac{5t^3}{3} + \cfrac{5t^2}{4} + 2t\right]_{0}^{10} \\&= \cfrac{5000}{3} + \cfrac{500}{4} + 20 \\&= 1811.667.. \ m\end{aligned} s = ∫ t 0 t v d t = ∫ t 0 = 0 s t = 10 s ( 5 t 2 + 2 5 t + 2 ) d t = [ 3 5 t 3 + 4 5 t 2 + 2 t ] 0 10 = 3 5000 + 4 500 + 20 = 1811.667.. m
Therefore, s = 1810 m ( 3 s . f . ) ‾ \underline{s = 1810 \ m \ (3 \ s.f.)} s = 1810 m ( 3 s . f . )
Example 2 A particle with m = 10 k g m=10 \ kg m = 10 k g x x x t t t s = ( 2 t 3 / 2 + e − 2 t 3 ) m s= \left(2t^{3/2} +\cfrac{ e^{-2t}}{3}\right) m s = ( 2 t 3/2 + 3 e − 2 t ) m
i) Calculate the velocity when t = 1.5 s t = 1.5 \ s t = 1.5 s
ii) Given that the particle is acted on by a force F F F , calculate the value of F F F when t = 10 s t=10 \ s t = 10 s to one decimal place.
i) Calculate the velocity when t = 1.5 s t=1.5 \ s t = 1.5 s to one decimal place.
Use v = d s d t v = \cfrac{ds}{dt} v = d t d s
v = d s d t = d d t ( 2 t 3 / 2 + e − 2 t 3 ) = 3 t − 2 e − 2 t 3 \begin{aligned}v &= \cfrac{ds}{dt} \\&= \cfrac{d}{dt}\left(2t^{3/2} + \cfrac{e^{-2t}}{3}\right)\\& = 3\sqrt{t} - \cfrac{2e^{-2t}}{3}\end{aligned} v = d t d s = d t d ( 2 t 3/2 + 3 e − 2 t ) = 3 t − 3 2 e − 2 t
Substitute t = 1.5 s t=1.5 \ s t = 1.5 s and you obtain:
v = 3.64 m s − 1 ( 3 s . f . ) ‾ \underline{v= 3.64 \ ms^{-1} \ (3 \ s.f.)} v = 3.64 m s − 1 ( 3 s . f . )
ii) Given that the particle is acted on by a force F F F , calculate the value of F F F when t = 10 s t=10 \ s t = 10 s to one decimal place.
By definition F = m × a F = m\times a F = m × a . Therefore, first you need to calculate the acceleration:
a = d v d t = d d t ( 3 t − 2 e − 2 t 3 ) = 4 e − 2 t 3 + 3 2 t \begin{aligned}a&=\cfrac{dv}{dt} \\&= \cfrac{d}{dt}\left(3\sqrt{t} - \cfrac{2e^{-2t}}{3}\right)\\&= \cfrac{4e^{-2t}}{3} + \cfrac{3}{2\sqrt{t}}\end{aligned} a = d t d v = d t d ( 3 t − 3 2 e − 2 t ) = 3 4 e − 2 t + 2 t 3
Substituting t = 10 s t = 10 \ s t = 10 s you have:
a = 0.47434165... m s − 2 a = 0.47434165... \ ms^{-2} a = 0.47434165... m s − 2
Thus, the force is:
F = 4.74 N ( 3 s . f . ) ‾ \underline{F = 4.74 \ N \ (3 \ s.f.)} F = 4.74 N ( 3 s . f . )
