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10. (a) A particle starts from rest and moves in a straight line with acceleration (25 − 10t) m s², where v is the speed of the particle - Leaving Cert Applied Maths - Question 10 - 2017
Question 10
10. (a) A particle starts from rest and moves in a straight line with acceleration (25 − 10t) m s², where v is the speed of the particle. (i) After time t, find v i... show full transcript
Worked Solution & Example Answer:10. (a) A particle starts from rest and moves in a straight line with acceleration (25 − 10t) m s², where v is the speed of the particle - Leaving Cert Applied Maths - Question 10 - 2017
Step 1
After time t, find v in terms of t.
Answer
To find the velocity v after time t, we start with the acceleration:
a = rac{dv}{dt} = 25 - 10t
Integrating both sides with respect to t:
rac{dv}{dt} &= 25 - 10t \\ ext{Integrating: } & ext{ } egin{align*} ext{ } \\ rac{dv}{dt} ext{ gives } \ \\ ext{ } v(t) &= rac{25t - 5t^2}{2} + C \\ ext{ Since it starts from rest: C=0; } \\ ext{ So, } & v(t) = rac{25t - 5t^2}{2}. \\ ext{At } t = 0, ext{ } v = 0. \\ ext{Thus, integrating gives: } \end{align*} ext{However, the solution directly from the integral is: } \end{align*} $$ v = 25 - 10e^{-0.1t} $$Step 2
Find the time taken to acquire a speed of 2.25 m s⁻¹ and find the distance travelled in this time.
Answer
To find the time taken to acquire a speed of 2.25 m s⁻¹:
Set
Rearranging gives:
e^{-0.1t} = rac{25 - 2.25}{10} = 0.2275.
Taking the natural logarithm:
t = - rac{ ext{ln}(0.2275)}{0.1} \\ t \approx 10.23 ext{s}.$$ Next, to find the distance travelled, using $$ ds = rac{1}{dt}v = 2.5(1 - e^{-0.1t}) \\ ext{Integrating from 0 to } t \text{ gives: } \\ ext{Distance} \approx 0.35 ext{m}.$Step 3
Step 4