Answer:

Explanation:
Given that the airplane starts from the rest (this is initial velocity equals to zero) and accelerates at a constant rate, position can be described like this:
where x is the position, t is the time a is the acceleration and
is initial velocity. In this way acceleration can be found.
.
Now we are able to found velocity at any time with the formula: 
Answer:
Volume of gasoline that expands and spills out is 1.33 ltr
Explanation:
As we know that when temperature of the liquid is increased then its volume will expand and it is given as

here we know that

volume expansion coefficient of the gasoline is given as

change in temperature is given as


Now we have


Answer:
d = 44.64 m
Explanation:
Given that,
Net force acting on the car, F = -8750 N
The mass of the car, m = 1250 kg
Initial speed of the car, u = 25 m/s
Final speed, v = 0 (it stops)
The formula for the net force is :
F = ma
a is acceleration of the car

Let d be the breaking distance. It can be calculated using third equation of motion as :

So, the required distance covered by the car is 44.64 m.
Choice-A is the correct one. It doesn't say it, but it means he'll see the ball reach the catcher's mitt first BEFORE HE HEARS IT slap the mitt.