Answer:four times
Explanation:
Given
mass of both cars A and B are same suppose m
but velocity of car B is same as of car A
Suppose velocity of car A is u
Velocity of car B is 2 u
A constant force is applied on both the cars such that they come to rest by travelling certain distance
using to find the distance traveled
where, v=final velocity
u=initial velocity
a=acceleration(offered by force)
s=displacement
final velocity is zero
For car A
[tex]0-(u)^2=2\times a\times s[/tex]
[tex]s_a=\frac{u^2}{2a}------1[/tex]
For car B
[tex]0-(2u)^2=2\times a\times s_b[/tex]
[tex]s_b=\frac{4u^2}{2a}----2[/tex]
divide 1 and 2 we get
[tex]\frac{s_a}{s_b}=\frac{1}{4}[/tex]
thus [tex]s_b=4\cdot s_a[/tex]
distance traveled by car B is four time of car A
By using the third equation of motion,
By substituting the values,
→ [tex]0 = (V_A^2)+2ad[/tex]
→ [tex]d = -\frac{V_A^2}{2a}[/tex]
and,
→ [tex]d_2 = -\frac{(V_B)^2}{2a} = -\frac{4 V_A ^2}{2a}[/tex]
hence,
→ [tex]d_B = 4d_A[/tex]
Thus the above response is appropriate.
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