A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).
We can calculate the initial velocity of the car (u) using the following kinematic equation.
![v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20u%5E%7B2%7D%20%2B%202as%5C%5C%5C%5Cu%20%3D%20%5Csqrt%5B%5D%7Bv%5E%7B2%7D-2as%7D%20%3D%20%5Csqrt%5B%5D%7B%280m%2Fs%29%5E%7B2%7D-2%28-42.61m%2Fs%5E%7B2%7D%20%29%2810.99m%29%7D%20%3D%2030.60m%2Fs)
A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
Learn more: brainly.com/question/14851168
Answer:
a = √ (a_t² + a_c²)
a_t = dv / dt
, a_c = v² / r
Explanation:
In a two-dimensional movement, the acceleration can have two components, one in each axis of the movement, so the acceleration can be written as the components of the acceleration in each axis.
a = aₓ i ^ + a_y j ^
Another very common way of expressing acceleration is by creating a reference system with a parallel axis and a perpendicular axis. The axis called parallel is in the radial direction and the perpendicular axis is perpendicular to the movement, therefore the acceleration remains
a = √ (a_t² + a_c²)
where the tangential acceleration is
a_t = dv / dt
the centripetal acceleration is
a_c = v² / r
Find Displacement and Distance
displacement ...
north is 700+400+100 =1200m n
south=1200m
1200-1200=0
east is 300+300=600m
west is 600m
600-600=0
back at dtart. displ zero
distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m = 3600m
Let the time be t.
so,In time t , distance travelled by 1st cyclist = 12.6 t
distance travelled by 2nd cyclist = 9.2t + 0.5 (1.5) t^2
Now, cyclist 1st is already 11.4 m ahead of 2nd cyclist.
so, 9.2t + 0.5 (1.5) t^2 = 11.4 + 12.6t
find t :
t = 6.77 sec
