A car has an initial speed of 25m/s and a constant deceleration of 3m/s^2. Determine the velocity of the car when t=4s. What is the displacement of the car during the 4-s time interval?
i do not know
what is the algebraic expression that'll help me solve this
the formula please
u should ask ur mom🙂✌️
a = -3 m/s^2
Vi = 25 m/s
final t = 4 s
===========
v = Vi - a t
x = Xi + Vi t - (1/2) a t^2
so
v = 25 - 3 * 4 = 25 - 12 = 13 m/s at t = 4
x - Xi = 25 *4 - (1/2) 3 *16
= 100 - 24
= 76 meters
or just average speed = (13 + 25) / 2
so distance = (13+25) *4/2 = 76 meters
To determine the velocity of the car when t=4s, 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
In this case, the initial velocity (u) is given as 25 m/s, the acceleration (a) is a deceleration of -3 m/s^2 (negative because it is slowing down), and the time (t) is 4s.
Let's plug in the values into the equation to find the velocity at t=4s:
v = u + at
v = 25 m/s + (-3 m/s^2)(4 s)
v = 25 m/s - 12 m/s
v = 13 m/s
So, when the time (t) is 4s, the velocity (v) of the car is 13 m/s.
To determine the displacement of the car during the 4-second time interval, we can use another equation of motion:
s = ut + (1/2)at^2
where:
s is the displacement
u is the initial velocity
t is the time
a is the acceleration
In this case, the initial velocity (u) is given as 25 m/s, the acceleration (a) is -3 m/s^2, and the time (t) is 4s.
Let's plug in the values into the equation to find the displacement during the 4-second time interval:
s = ut + (1/2)at^2
s = (25 m/s)(4 s) + (1/2)(-3 m/s^2)(4 s)^2
s = 100 m + (1/2)(-3 m/s^2)(16 s^2)
s = 100 m + (-24 m)
s = 76 m
So, during the 4-second time interval, the car traveled a displacement of 76 meters.