A car moving with constant acceleration covers the distance between two points 180m apart in 6sec.its speed as it passes the second point is 45m/s.what is its acceleration and its speed at the first point.
measuring from the first point,
s(t) = vt + 1/2 at^2
we know that s(6) = 180, so
6v + 18a = 180
Also, we have
v + 6a = 45
So, v=15 and a = 5
Acc. to the question v=u+at ;v=45m/s , t=6s ;45=u+6a ;(45-u)/6=a----1 ; v^2-u^2=2as ; 2025-u^2=360a ---2 ; subsitute --1 in ---2 eq ; 2025-u^2=360*(45-u)/6 by solving we will get a equation ; u^2-u+675 ;we will get u=15 or 45 ; we have to take 15 =u; then v=u+at ; 45=15+6a ; a=5m/s
Acc. to the question v=u+at ;v=45m/s , t=6s ;45=u+6a ;(45-u)/6=a----1 ; v^2-u^2=2as ; 2025-u^2=360a ---2 ; subsitute --1 in ---2 eq ; 2025-u^2=360*(45-u)/6 by solving we will get a equation ; u^2-u+675=0 ;we will get u=15 or 45 ; we have to take 15 =u; then v=u+at ; 45=15+6a ; a=5m/s
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To find the acceleration of the car, we can use the equation of motion:
\[ v = u + at \]
where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
Given:
- Distance covered, d = 180 m
- Time, t = 6 s
- Final velocity, v = 45 m/s
First, let's calculate the acceleration:
Step 1: Calculate the initial velocity (u)
We know that the initial velocity is zero because the car starts from rest at the first point.
Therefore, u = 0 m/s.
Step 2: Calculate the acceleration (a)
Using the equation mentioned above, we can rearrange it as follows:
\[ a = \frac{{v - u}}{{t}} \]
Substituting the known values:
\[ a = \frac{{45 - 0}}{{6}} = 7.5 \, \text{m/s}^2 \]
The car's acceleration is 7.5 m/s².
Now, let's calculate the car's speed at the first point.
Using the equation of motion again:
\[ v = u + at \]
This time, we want to find the initial velocity (u). So, we rearrange the equation:
\[ u = v - at \]
Substituting the known values:
\[ u = 45 - (7.5 \times 6) = 0 \, \text{m/s} \]
The car's speed at the first point is 0 m/s (because it starts from rest).