change in speed ?

Question

a scateboarder rolls down a hill with the acceleration of 4.0m/s2(squared). he is on a hill for 4.8s (seconds). what is the change in his speed ?

m-meters s-seconds
explain in full detail formula

Answer 1

acceleration is defined as the change in speed per unit time.

the skateboarder accelerates at 4.0 m s-2 for 4.8s

change in speed = acceleration*time {definition}
change in speed = 4.0*4.8
change in speed = 19.2 m/s

Answer 2

Acceleration by definition is
a = dv/dt
So the speed v(t) at any time, t, is
v(t) = the integral of 'a' with respect to 't'.
When 'a' is constant, its simply
v(t) = a*t + C , equation 1
( C is the arbitrary constant of integration)
Since v(0) = a*0 + C (the speed at t=0)
C = v_0 (initial speed or v(0) )
From equation 1,
v(t) = a*t + v_0
So the final speed v_f , ( v at t = t_f or v(t_f) )
v_f = a*t_f + v_0
Therefore the change in speed
v_f - v_0 = a*t_f
= 4*4.8 = 19.2 m/s

Answer 3

Hi,
The speed or velocity (speed has no direction, velocity does) is the initial velocity (call it Vo) + the acceleration (a) multiplied by the time (t). Translated to algebra, that's this:
V = Vo + at
So, in your case apparently Vo = 0, so we have this:
v = at
= 4.0*4.8
= 19.2 m/s

Hope this helps.
FE

Answer 4

lets take these
initial velocity = u = 0
final velocity = v = ?
time = t = 4.8
acceleration = a = 4m/s2
by the eqn of motion v = u +at
v = 0 + 4 x 4.8
v = 19.2
so the cange in speed is v - u
19.2 - 0
19.2m/s

clear?

Answer 5

s=d/t
83.3m/s