A car, initially travelling at the speed 100 km/h, slows down according to
the formula
L(t) = At − Bt2.
where L is the travelled distance, t is the time, and B = 90 km/h2. Using
derivative, find the time moment when the car speed becomes 10 km/h. Find
the acceleration of the car at this moment.
+33661 Speed is the rate of change of distance with time. Differentiating L(t) = At - Bt2 with respect to time (t) we get speed, S as
S(t) = dL(t)/dt = A - 2Bt
When t = 0, S(0) = 100km/hr, so:
100 = A - 2B*0
or A = 100 km/hr.
when S(t) = 10 km/hr we have
10 = 100 - 2*90*τ (where I've used τ for the time at which speed is 10 km/hr).
Rearranging this we get
τ = (100 - 10)/(2*90) = 1/2
so τ = 1/2 an hour or 30 minutes. (That's a long time to slow down!!).
+33661 Speed is the rate of change of distance with time. Differentiating L(t) = At - Bt2 with respect to time (t) we get speed, S as
S(t) = dL(t)/dt = A - 2Bt
When t = 0, S(0) = 100km/hr, so:
100 = A - 2B*0
or A = 100 km/hr.
when S(t) = 10 km/hr we have
10 = 100 - 2*90*τ (where I've used τ for the time at which speed is 10 km/hr).
Rearranging this we get
τ = (100 - 10)/(2*90) = 1/2
so τ = 1/2 an hour or 30 minutes. (That's a long time to slow down!!).
