Question

1. The velocity of a car as it travels over a level crossing is modeled by v = t2 – 4t + 12

Answer 1

Answer:

At t=2s the car  reach its minimum speed

Minimum speed=8m/s

Step-by-step explanation:

We are given that velocity of car

$v(t)=t^2-4t+12$

0<t<4

Differentiate w.r.t t

$v'(t)=2t-4$

$v'(t)=0$

$2t-4=0$

$2t=4$

$t=2$

$v''(t)=2>0$

The velocity is minimum at t=2 s

Substitute t=2

$v(2)=2^2-4(2)+12$

$v(2)=8m/s$

Hence, the speed of car is minimum at t=2 se and minimum speed of car=8m/s