We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as
Distance = speed x time
From the information given,
speed = 320 m/s
time = 4.5 s
By substituting these values into the formula, we have
Distance = 320 m/s x 4.5s
s cancels out. We are left with m. Thus,
Distance = 1440m
Answer:
v_average = (d₂-d₁) / Δt
this average velocity is not necessarily the velocity of the extreme points,
Explanation:
To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.
v_average = (d₂-d₁) / Δt
this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.
Therefore neither of them is right.
The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².
Answer:
Explanation:
Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.
Then acceleration = change in velocity/Time.

Acceleration = (9-0)/3=9/3=3 m/s².
So the cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s