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

with 5 seconds, no, but over time if the police car stays at 65m/s ad the speeding car stays at 60, then obviously yes, but the answer to your question would be, no.

hope this : )