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


For car B


divide 1 and 2 we get

thus 
distance traveled by car B is four time of car A
Answer:
30 m/s
Explanation:
Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.
Answer:
6.53 m/s²
Explanation:
Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.
Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:
T = m₁a (1)
m₂g - T = m₂a (2)
substituting T = m₁a in equation 2:
m₂g - m₁a = m₂a
m₂a + m₁a = m₂g
a(m₁ + m₂) = m₂g
a = m₂g / (m₁ + m₂)
a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²
Both objects have an acceleration of 6.53 m/s²