How long (in seconds) does it take a car accelerating at 3.1 m/s2 to go from rest to 51 m/s?
1 answer:
Answer:
t = 16.5s
Explanation:
Given parameters:
Acceleration = 3.1m/s²
Initial velocity = 0m/s
Final velocity = 51m/s
Unknown:
Time taken = ?
Solution:
To solve this problem we need to reiterate that acceleration is the rate of change of velocity with time.
So;
Acceleration =
v is the final velocity
u is the initial velocity
t is the time taken
So;
3.1 =
3.1t = 51
t = 16.5s
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Answer:
Explanation:
Yes , the horizontal distance travelled by the ball will depend upon the height of release .
When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play . But the horizontal range covered by the body thrown
depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .
Horizontal distance covered = t x horizontal velocity
If V be the velocity of throw and Vx be its horizontal component
Horizontal distance covered = t x Vx
Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .
If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity
from the relation
s = ut + 1/2 at²
h = - Vy t + 1/2 at²
As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only horizontal velocity initially , Vy = 0
h = 1/2 gt²
Horizontal distance covered = t x Vx
=
From this expression also
Horizontal distance covered is proportional to .