Bad drivers

The hard way to do this is to stay in the reference frame all the velocities are quoted in, equate the final position and velocity of your car with the truck (as that's what you need to avoid rear-ending the truck), and solve the resulting equations. An easier way is to switch to the reference frame of the truck. Since this is an inertial reference frame (the truck moves at a constant velocity) the acceleration we calculate in this frame is the same. In the truck's frame the initial velocity of your car is 10 m/s and it is 20 m behind you. It then reaches a velocity of 0 m/s when it almost touches the truck. Therefore we have,

$v_f^2=v_i^2+2ad \rightarrow 0^2=10^2+2a(20)$

which allows us to immediately solve for $a=-2.5~m/s^2$ .

Let the desired acceleration be $a$ . Then, the time we care about is $t=\frac{v_f-v_i} {a}=-\frac{10}{a}$ . This is the time it takes for you to decelerate to $20$ m/s. Since the velocity is always positive, we only have to make sure that we won't rear-end the front car at $-\frac{10}{a}$ seconds, or the maximal bound. Therefore, we have $30t-\frac{1}{2}a{t^2}=20+20t$ . Plugging in $t=\frac{10}{a}$ gives $\frac{100}{a}-\frac{50}{a}=20$ . Therefore, our braking acceleration is $\boxed{-2.5}$ m/s.

we have Va wrt. b=Va-Vb. now U minecar wrt.V bad driver's car=30-20 =10m/s DISPLACEMENTminecar wrt. bad driver's care=20m

final velocity; Vminecar wrt.V bad driver's car=0 m/s (since we have to stop relative to next car to prevent accident)

let accleration of my car be ''a m/s^2''

we have U=10m/s V=0m/s S=20m ACCL.=a m/s^2

using newon's third equation of motion with constant accleration V^2=U^2+2AS we get' a= -100/40 = -2.5 m/s^2

The relative velocity is 30 - 20 =10 m/s. Acceleration is constant. End velocity is 0 m/s. So the average velocity is 5 m/s. So the time taken is (20 m)/ (5 m/s) = 4 sec. To destroy the relative velocity of 10 m/s in 4 sec,
acceleration = (10 m/s)/ 4sec =2.5 m/sec/sec. $\\~\\\boxed{2.5}$

Equation for Distance for Orange Truck (Driver in Front) = $d = 20 \cdot t + 20$ Equation for Distance of Your Car = d = 30 \cdot t + .5 \cdot a \cdot t^2 The key is realizing that at the instant the vehicles are at the same distance, the velocity of your car must be 20. If it is greater than 20, you will rear end the car. With this in mind: 20 = 30 + a \cdot t Or a \cdot t = -10 Keep in mind: In this last equation, the "t" variable represents the time at which the vehicles are at the same position.

To find the t such that the vehicles are at the same position, set the distance equations equal to each other:

20 \cdot t + 20 = 30 \cdot t + .5 \cdot a \cdot t^2 After some simple algebra, this simplifies to: 20 = 10 \cdot t + .5 \cdot (a \cdot t) \cdot t Remembering that we previously solved that a \cdot t = -10, we substitute that into our equation to get: 20 = 10 \cdot t + .5 \cdot (-10) \cdot t Or: 20 = 10 \cdot t - 5 \cdot t = 5 \cdot t Dividing by 5, we find that t = 4. Since a \cdot t = -10, a = -2.5, and this is the answer.