Objects A and B both start from rest. They both accelerate at the same rate. However, objects A accelerate for

Question

twice the time as object B. What is the distance traveled by object A compared to that of object B?

Answer 1

v(t)=a*t
(Since there is no initial velocity, the vi component is zero)
d(t)=.5*a*t^2

dB=.5*a*t^2
dA=.5*a*4*t^2

where t is the time of acceleration

dA/dB=4

for times greater than the acceleration time, the equations will be different for distance. If you need those, just post a comment.

j

Answer 2

Both objects position is determined by the equation:

s = 1/2 a t^2
with s being position, a being acceleration, and t being time.

If object A accelerates twice as long as object B, then:
t_1 = 2* t_2

Substitute 2 * t_2 in place of t in object A's equation and you get:

s = 1/2 a (2 t)^2

2 squared equals 4, so object A travels four times as far as object B

Answer 3

If we calculate the speeds in line with variables x=acceleration and t=time then we can locate the assessment velocity of b=xt velocity of a=x(2t)=2xt as you will discover, the fee of a is two times that of b if the time is accelerated by using 2

Answer 4

x = vt + 1/2at^2

because they start from rest.

x= 1/2at^2

do that for each object and find the difference