Answer to Question #101492 in Calculus for Joe

Question

The car's speed after braking is v (t) = 16.9 - 0.1t2 (where time t is seconds).

a) How long does the braking last?

(b) the distance traveled by the first 3 and last 3 cars
within seconds of starting braking?

c) How long does the car move during the entire braking?

Answer 1

a) The braking process occurs until the moment when the speed becomes equal to zero. Hence

$$16.9-0.1t^2=0 \implies t^2=169 \implies t=13s$$

b) We find the required distance using the formula

$$d=\smallint_0^{3}v(t)dt+\smallint_{10}^{13}v(t)dt=\smallint_0^{3}(16.9-0.1t^2)dt+\smallint_{10}^{13}(16.9-0.1t^2)dt=$$

$$=(16.9t-\frac{0.1t^3}{3})_0^{3}+(16.9t-\frac{0.1t^3}{3})_{10}^{13}=16.9\times3-\frac{0.1\times{3^3}}{3}-0+$$

$$+16.9\times13-\frac{0.1\times{13^3}}{3}-16.9\times10+\frac{0.1\times{10^3}}{3}=60.6$$

c) During braking, the car travels a distance equal to

$$d=\smallint_0^{13}v(t)dt=\smallint_0^{13}(16.9-0.1t^2)dt=(16.9t-\frac{0.1t^3}{3})_0^{13}$$ $$d=16.9\times13-\frac{0.1\times{13^3}}{3}-0=146.47$$

Answer. a) 13s. b) 60.6 c) 146.47.