Answer:
The position at time t is [tex]s(t) = 8t^2 - 14t -8[/tex]
Step-by-step explanation:
From the question we are told that
The acceleration is [tex]a = 16[/tex]
The velocity at t = 0 is [tex]v(0) = -14[/tex]
The position at time t = 0 is [tex]s(0) = -8[/tex]
Generally acceleration is mathematically represented as
[tex]a(t) = \frac{d v}{dt }[/tex]
=> [tex]\frac{dv}{dt} = 16[/tex]
=> [tex]dv = 16 dt[/tex]
integrating both sides we have
[tex]\int\limits dv = \int\limits16 dt[/tex]
=> [tex]v(t) = 16 t + c[/tex]
Now at t = 0
[tex]v(0) = 16 * 0 +c = -14[/tex]
=> [tex]c = -14[/tex]
So
[tex]v(t) = 16 t -14[/tex]
Generally the position of the body is mathematically represented as
[tex]s(t) = \int\limits v(t)dt[/tex]
So
[tex]s(t) = \int\limits 16t - 14 dt[/tex]
So
[tex]s(t) = 16 \frac{t^2}{2} - 14t + C[/tex]
Now at t = 0
[tex]s(0) = 16 \frac{0^2}{2} - 14(0) + C = -8[/tex]
=> [tex]C = -8[/tex]
So
[tex]s(t) = 8t^2 - 14t -8[/tex]
Rate of change of velocity with respect to time is known as acceleration.
Rate of change of displacement with respect to time is known as velocity.
Body position at time t is given by, [tex]s(t)=8t^{2}-14t-8[/tex]
Data given are, a=16, v(0) = -14, s(0) = -8
Since, [tex]\frac{dv}{dt}=a\\\\\frac{dv}{dt}=16\\\\dv=16dt[/tex]
Integrating both side
We get, [tex]v=16t + C[/tex]
Substituting v(0)=-14 in above equation
We get, C = -14
So, [tex]v=16t-14[/tex]
Since, Rate of change of displacement with respect to time is known as velocity.
[tex]v=\frac{ds}{dt}\\\\ds=vdt\\\\ds=(16t-14)dt[/tex]
Taking integration on both side
We get, [tex]s=16\frac{t^{2} }{2}-14t+c[/tex]
Substituting s(0)= - 8 in above equation
We get, [tex]c=-8[/tex]
So, [tex]s=16\frac{t^{2} }{2}-14t-8\\\\s(t)=8t^{2} -14t-8\\[/tex]
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