Re-arrange the equation x = vit + 1/2at2 v2f = vi2 + 2ax vf = vi + at x = 1/2(vi + vf)delta t
1. A car in front of the school goes from rest to 27 ms in 3.0 seconds. What is its acceleration (assuming it is constant)?
2. A rocket starts from rest and accelerates at a uniform rate of 190 m/s^2 for 2.4 seconds. What is its final velocity?
3. A car has a velocity of 15 m/s. It then accelerates at a uniform rate of 3.5 m/s^2 for the next 5.0 seconds. What distance does the car cover during this time?
4. What distance is required for a train to stop if its initial velocity is 23 m/s and its deceleration is 0.25 m/s^2? (Assume the train decelerates at a constant rate.
5. What distance will a car cover while uniformly accelerating from 12 m/s to 26 m/s in 14 seconds?
6. A person starts at rest and accelerates at a uniform rate of at 3.2 m/s^2 for 3.0 seconds. What distance does this person cover in that time?
7. What must be the acceleration of a train in order for it to stop from 12 m/s in a distance of 541 m? (Assume the train accelerates uniformly.)

Explanation:

1) Acceleration is the change in velocity of a body with respect to time.

Acceleration a = vf - vi/t

vf is the final velocity

vi is the initial velocity.

t is the time taken

Since the body accelerates from rest, vi = 0m/s

a = 27-0/3

a = 27/3

a = 9m/s²

2) Given

u = 0m/s (accelerates from rest)

a = 190m/s²

t = 2.4seconds

v = ?

Using v = u+at

v = 0+190(2.4)

v = 190×2.4

v = 456m/s

The final velocity of the car is 456m/s

3) Given

u = 15m/s

a = 3.5m/s

t = 5seconds

Using the relationship

S = ut+1/2at²

S is the distance covered by the car.

S = 15(5)+1/2(3.5)×5²

S = 75+25×3.5/2

S = 75+43.75

S = 118.75m

4) Given

initial velocity u = 23m/s

Deceleration a = -0.25m/s²(negative acceleration)

Final velocity v = 0m/s

Using the relationship

V² = u²+2as

0² = 23²+2(-0.25)s

-23² = -0.5S

23² = 0.5S

S = 529/0.5

S = 1058m

The distance required for the train to stop is 1058m.

5) Given

initial velocity u = 12m/s

Final velocity v = 26m/s

time = 14sec

Acceleration a = v-u/t

a = 26-12/14

a= 14/14

a = 1m/s²

For Distance covered

v² = u²+2as

26² = 12²+2(1)S

676 = 144 +2S

2S = 676-144

2S = 532

S = 532/2

S = 266m

Distance that the car will cover is 266m

6) Given

Initial velocity u = 0m/s (person starts from rest)

acceleration a = 3.2m/s²

time t = 3.0s

To get the distance;

S = ut + 1/2at²

S = 0(3)+1/2(3.2)×3²

S = 0+1.6×9

S= 9×1.6

S = 14.4m

The distance that the person covered in 3.0s is 14.4m

7) initial velocity of train u = 12m/s

Distance covered S = 541m

Final velocity = 0m/s (on stopping)

acceleration a= ?

Acceleration will be negative since the train is coming to a stop (decelerating)

Using the formula v² = u²+2as

0² = 12² - 2a(541)

-12² = -1082a

144 = 1082a

a = 1082/144

a = 7.51m/s²

Hence the acceleration of a train in order for it to stop from 12 m/s in a distance of 541 m is 7.51m/s²

answer: the direction of the wind is expressed as the direction from which the wind is blowing. for example, easterly winds blow from east to west, while westerly winds blow from west to east.

explanation:

a person with a mass of 106kg has a weight of 1060n