Answer:
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
