Answer:
Explanation:
The friction created between the tire and the ground generates thermal energy as force is applied during skidding.
The mentioned force relates to half the impact on the rear tire, resulting in a calculated normal force of,
The work executed is determined by the frictional force and the distance covered,
Where
The thermal energy produced from the work done is,
Answer:
The distance before stopping is 1.52 m,
velocity is 4.0 m/s on the y-axis
Explanation:
The particle’s motion is two-dimensional due to acceleration along both the x and y axes; each axis can be addressed independently for calculations.
a) At the moment the particle starts to reverse, its velocity should be zero (Vfx = 0)
Vfₓ = V₀ₓ + aₓ t
t = - V₀ₓ/aₓ
t = - 2.4/(-1.9)
t= 1.26 s
When the particle stops, we calculate its position
X1 = V₀ₓ t + ½ aₓ t²
X1= 2.4 1.26 + ½ (-1.9) 1.26²
X1= 1.52 m
At this point, the particle begins returning.
b) The velocity comprises both x and y components.
For the x section, Vₓ = 0 m/s indicates a halt, but the y component retains a velocity
Vfy= Voy + ay t
Vfy= 0 + 3.2 1.26
Vfy = 4.0 m/s
Thus, the velocity reads as
V = (0 x^ + 4.0 y^) m/s
c) To graph the motion, we create a table listing position x and y at given time intervals; let's begin the calculations for equations
X = V₀ₓ t+ ½ aₓ t²
Y = Voy t + ½ ay t²
X= 2.4 t + ½ (-1.9) t²
Y= 0 + ½ 3.2 t²
X= 2.4 t – 0.95 t²
Y= 1.6 t²
With these equations, we construct two graphs for position against time, one for the x-axis and another for the y-axis
Chart for graphing
Time (s) x(m) y(m)
0 0 0
0.5 0.960 0.4
1 1.45 1.6
1.50 1.46 3.6
2.00 1.00 6.4
1) 9.18 s
During the initial phase, the rocket accelerates at
over a time interval of
The final velocity after this period is found using the SUVAT formula:
with initial velocity u = 0. Substituting a1 and t1 gives:
Afterward, the rocket decelerates uniformly at
until it stops, meaning the final velocity is
Again using the SUVAT formula,
and knowing , solve for t2, the time until the rocket halts:
2) 299.9 m
Calculate distances covered in both phases.
Distance in the first phase:
Substituting values from part 1,
Distance in the deceleration phase:
Substituting known values,
Total distance traveled is
d = 82.7 m + 217.2 m = 299.9 m