I hope that the attachment helps you..
1.5 m/s is the velocity.
9.3 m is the length of aisle, over which Distance will be covered.
Time is demanded in which the child will move the cart over the aisle with 1.5 m/s.
v=S/t
and,
t=S/v
Put values,
t=9.3/1.5=6.2 s
Answer:
45 s .
Explanation:
The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .
displacement s = ?
acceleration a = 1 m /s²
Final speed v = 5 m/s
initial speed u = 0
v² = u² + 2as
5² = 0 + 2 x 1 x s
s = 12.5 m
B) Let time of acceleration or deceleration be t
v = u + a t
5 = 0 + 1 t
t = 5 s
Similarly displacement during deceleration = 12.5 m
Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) = 175 m .
velocity of uniform motion = 5 m /s
time during which there was uniform velocity = 175 / 5 = 35 s
Total time = 5 + 35 + 5 = 45 s .
Answer:
The force of the impact would be smaller
Explanation: Examples:
If the force is big then the time would be small (2500N of Force = 10 seconds)
If the force is small then the time would be big (250N of Force = 50 seconds)
Impulse/Collision -> [Ft] = [M (vf-vo)] <- Change in momentum
