Displacement word problem grade 11

[max1020]

1. while driving a lady sees a cat run on the road and it takes her 0.80s to react and put her foot on the brake. her car is traveling 25 m/s. (a)how far will the car travel before she puts her foot on the brake and (b) if when the brake is applied the car decelerates at a uniform rate of 9.3 m/s^2, what is the total displacements in the car?





i tried dividing .80s by 25 m/s for (a) but i don't know if its right. and i have no idea how to do the (b)

 

[jedishrfu]

You take one step per second and you walk for 3 seconds how many steps do you take?

The lady is traveling at 25 meters/second and travels for 0.8 of a second before hitting the brake and how many meters does she travel?

for part b, show us what formulas you know that might apply to the problem and think I have time, speed and acceleration and want distance...

 

[max1020]

You take one step per second and you walk for 3 seconds how many steps do you take?

The lady is traveling at 25 meters/second and travels for 0.8 of a second before hitting the brake and how many meters does she travel?

for part b, show us what formulas you know that might apply to the problem and think I have time, speed and acceleration and want distance...

So I would multiply .80*25 which is 20m

 

[jedishrfu]

So I would multiply .80*25 which is 20m

Yes and you can check via units of measure m/s *s is meters.

 

[HalfLostAndSearching]

part

For part (a), your approach is correct. To find the distance the car travels before the lady puts her foot on the brake, we can use the formula d = vt, where d is the distance, v is the velocity, and t is the time. In this case, v = 25 m/s and t = 0.80 s. So, the distance traveled before the lady reacts and applies the brake is:

d = (25 m/s) x (0.80 s) = 20 meters

For part (b), we can use the formula d = (v0)t + (1/2)at^2, where d is the displacement, v0 is the initial velocity, a is the acceleration, and t is the time. In this case, v0 = 25 m/s, a = -9.3 m/s^2 (negative because it is decelerating), and t = 0.80 s. So, the total displacement of the car can be calculated as:

d = (25 m/s)(0.80 s) + (1/2)(-9.3 m/s^2)(0.80 s)^2 = 20 meters - 2.368 meters = 17.632 meters

Therefore, the total displacement of the car is 17.632 meters. It is important to note that displacement is a vector quantity, meaning it has both magnitude and direction. In this problem, the direction of the displacement is opposite to the direction of the car's initial velocity.