Question

Physics help... there a few questions I need help with?

1. A plane is travelling at an average speed of 891 km/h. How far does the plane travel (km) in 54.4 minutes?

2. What is the average speed (km/h) of a sprinter who runs 11.7 meters in 16.4 seconds?

3. A parked police car takes off after a suspect and reaches 80.2 km/h E in 8.16 seconds. What is the average acceleration (m/s2) of the car?

4.While driving south on hwy 400, Jim gets stuck travelling at 1.89 m/s behind a slow truck. He finally gets into a passing lane, hits the gas and speeds up with a constant acceleration of 0.0570 m/s2. What is his velocity (m/s) after accelerating for 8.49km?

5.How long (s) will it take a car travelling SE to traverse a distance of 719 m if it accelerates from rest at a rate of 2.1 m/s2 ?

6.Jenny throws a ball straight down from the balcony of an apartment building at a velocity of 8.38 m/s. Calculate the velocity (m/s) of the ball after 1.12 s. Neglect air resistance.

7.A basketball is thrown vertically upward with an initial velocity of 22.8 m/s. 'a) How long (s) does the ball take to reach its maximum height?
'b) How high (km) does the basketball go?

1. In the sketch below, a) What is the x-component of vector B? b) What is the y-component of vector B?

Answer 1

1. average speed = total distance /total time

given, average speed = `891*(5/18) m/s` and time =`54.5* 60 s`

So,distance covered = `809325 m`

2 . using the same formula,average speed =`11.7/16.4 m/s = 0.71 m/s`

3 . If, the police reaches a velocity of `80.2 *(5/18) m/s` from `0 m/s` in `8.16 s` then average acceleration = change in velocity/time required= `2.73 m/s^2`

4 . As he went for accelerating constantly,we can apply `v^2 = u^2 +2gs` (where, `v` is the velocity after going through distance `s` with an initial velocity `u`)

Putting the given values we get,`v=31.17 m/s`

5 . Here we can apply `s=1/2 at^2`(where, `s` is the displacement after time `t` with an acceleration `a`)

Putting the given values we get, `t=26.17 s`

6 . Here for free fall we can apply `v=u+g t` (where, `v` is the velocity of a ball falling under gravitational acceleration `g` that had an initial velocity of `u`)

Putting the given values we get, `v=8.38+ 9.8*1.12 =19.356 m/s`

7 . here,we can use `v=u-g t` (expression is same as before,only here velocity has the direction against `g`)

At the highest point of its motion the basket ball will loose its velocity,so `v` will become `0` and for this it will take time `t`,

given,`u=22/8 m/s`

So, `t =2.32s`

And, using `v^2 = u^2 -2gs` ,we can get the maximum height,

So, `s=26.52 m`

8 . X component of `vec B` is `21.5 sin 52=16.94 N` and y component is `21.5 cos 52=13.23 N`