The driver of a pickup truck going 89.9 km/h applies the brakes giving the truck a uniform deceleration of 5.99 m/s^2 while it travels 19.3 m. a. the velocity of the truck in m/s at the end of this displacement is m/s. b. how much time has elapsed?

(a) First, convert the truck's velocity from km/h to m/s:

Then the velocity after traveling satisfies

where and are the truck's initial velocity and acceleration, respectively. So we have

Taking the square root gives two possible velocities of the same magnitude but in opposite directions. It's technically feasible that over the given displacement, the truck changes its direction and is accelerating back towards its starting position since we're treating the truck as a particle.

(b) When acceleration is constant, average velocity is given by

which tells us that

Plugging in both possible values of we found earlier, we get (if is positive) or (if is negative).

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

this is a good question! i'll do some research and come back when i have the answer!