Question

At a certain moment, a car travelling at 40mph begins braking for a railroad crossing. Assume thatthe acceleration is constant (and negative!) and that 2 minutes later, the car comes to a stop, justbefore the crossing.(a) Find a formula for the velocity v(t) at any moment during the deceleration.(b) Determine how far away from the railroad crossing the car was when it began decelerating.

Answer 1

Answer:

Part a)

$v_f = 17.88 - (0.15) t$

Part b)

$d = 1072.8 m$

Explanation:

As we know that initial speed of the car is

$v = 40 mph$

$v = 40 \times (\frac{1609}{3600})$

$v = 17.88 m/s$

now we have

$v_f = v_i + at$

$0 = 17.88 + a(120)$

$a = -0.15 m/s^2$

Part a)

As we know that acceleration is constant here

so we have

$v_f = v_i + at$

$v_f = 17.88 - (0.15) t$

Part b)

distance moved by the car is given as

$d = \frac{v_F + v_i}{2} t$

now we have

$d = \frac{0 + 17.88}{2}(120)$

$d = 1072.8 m$