Question

wo cars are traveling along a straight line in the same direction, the lead car at 25.7m/s and the other car at 29.2m/s. At the moment the cars are 39.7m apart, the lead driver applies the brakes, causing her car to have an acceleration of -1.99m/s2. How long does it take for the lead car to stop? 1.29×101 s right

What is the distance it travels during this time? 1.66×102 m right

Assuming that the chasing car brakes at the same time as the lead car, what must be the chasing car's minimum negative acceleration so as not to hit the lead car?

How long does it take for the chasing car to stop?

Relative mohion aAg aA-aer@a-.,.91).m)s2. = mow

the picture below is wrong , need the right calculations for this please.

Answer 1

Let the chasing car has an acceleration a in backward direction.

Leading car = car A

chasing car = car B

relative speed = V = V_B - V_A = 29.2 - 25.7 = 3.5 m/sec

realtive distnace = 39.7 m

relative acceleration = -a -(- 1.99) = 1.99 -a

to prevent the collision relative motion should stop before the relative distance is covered:

v² = u² + 2as

0 = (3.5)² + 2*(1.99-a)*39.7

1.99 - a = - 0.154

so minimum acceleration of chasing car = 2.144 or 2.15 m/sec² backward

time need to stop :

using v = u +at

0 = 29.2 - 2.15t

t = 13.58 sec