Answer:
increase.
Explanation:
According to the newton’s second law of motion force is expressed as product of mass and acceleration.
F = m a
If the force acting is constant, then.
m∝ 
That is if the mass of object increases the acceleration decreases and vice versa. The above equation is used when the force acting on the body is constant.
As the thrust force from the rocket engine is constant throughout there will be a variation in the mass or acceleration.
Thus, it won't stay the same.
As the weight of the car is maximum at the start because of the fuel present in the rocket engine and minimum at the end as the fuel burns throughout the journey of the car. Weight will be minimum at the end and hence acceleration is maximum at the end.
Thus, it won't decrease.
As the acceleration is going from minimum at the start to maximum at the end, therefore it is continuously increases throughout its journey.
Thus, it will increase.
Answer:
solution:
to find the speed of a jogger use the following relation:
V
=
d
x
/d
t
=
7.5
×m
i
/
h
r
...........................(
1
)
in Above equation in x and t. Separating the variables and integrating,
∫
d
x
/7.5
×=
∫
d
t
+
C
or
−
4.7619
=
t
+
C
Here C =constant of integration.
x
=
0 at t
=
0
, we get: C
=
−
4.7619
now we have the relation to find the position and time for the jogger as:
−
4.7619 =
t
−
4.7619
.
.
.
.
.
.
.
.
.
(
2
)
Here
x is measured in miles and t in hours.
(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),
to get:
= −
4.7619
=
1
−
4.7619
= −
3.7619
or x
=
7.15
m
i
l
e
s
(b) To find the jogger's acceleration in m
i
l
/
differentiate
equation (1) with respect to time.
we have to eliminate x from the equation (1) using equation (2).
Eliminating x we get:
v
=
7.5×
Now differentiating above equation w.r.t time we get:
a
=
d
v/
d
t
=
−
0.675
/
At
t
=
0
the joggers acceleration is :
a
=
−
0.675
m
i
l
/
=
−
4.34
×
f
t
/
(c) required time for the jogger to run 6 miles is obtained by setting
x
=
6 in equation (2). We get:
−
4.7619
(
1
−
(
0.04
×
6 )
)^
7
/
10=
t
−
4.7619
or
t
=
0.832
h
r
s
To find speed you have to divide distance by time. In this case:
5 meters➗3 seconds = about 1.66666666 and so on m/s.
You could round to 1.67 or 1.7 if you'd like.
Part a
Answer: NO
We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.
Using the second equation of motion:
where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.
It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, 
Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.
Part b
Answer: 29.6 m/s
The maximum distance that car can travel is 
The acceleration is same,
The final velocity, v=0
Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:
Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.
