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Andrei [34K]
3 years ago
10

In a football game, the running back takes a handoff and begins running toward midfield at 3.21 yards/s . As he moves through hi

s linemen, he accelerates at a rate of 1.71 yards/s^2 to a velocity of 7.54 yards/s before he is tackled. How far did he run after he got the ball?
Physics
1 answer:
Inessa [10]3 years ago
8 0

Given Information:

Initial speed = u = 3.21 yards/s

Acceleration = α  = 1.71 yards/s²

Final speed = v = 7.54 yards/s

Required Information:

Distance = s = ?

Answer:

Distance = s = 13.61

Explanation:

We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.

We know from the equations of motion,

v² = u² + 2αs

Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.

Re-arranging the above equation for distance yields,

2αs = v² - u²

s = (v² - u²)/2α

s = (7.54² - 3.21²)/2×1.71

s = 46.55/3.42

s = 13.61 yards

Therefore, the runner traveled a distance of 13.61 yards before being tackled.

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a trolley of mass 20kg was originally at rest on a smoth horizontal surface. bu how much will it accelerate if a pulling force o
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Answer:

just divide 22 N by 20 kg to get the acceleration in m/s2

Explanation:

I hope this is right-

6 0
3 years ago
As you sit in a chair, in what direction (or directions) do you exert an action force?
enyata [817]
As we sit in a chair, Action force will be only in one direction and that direction would be downward only.

In short, Your Answer would be Option A

Hope this helps!
5 0
4 years ago
Read 2 more answers
A car accelerates from 13 m/s to 25 m/s in 5.0 s. assume constant acceleration. what was its acceleration?
natima [27]
<span>a = 25-13/6  = 12/6 = 2 m/s^2
Av speed: 25+13/2 = 38/2  = 19 m/sec
Dist = speed * time
19 * 6 = 114 meters</span>
8 0
3 years ago
Two postal delivery workers have different routes. They both travel from the post office, to neighborhoods to deliver mail, and
SOVA2 [1]

they travel the same distance

7 0
3 years ago
In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the ancho
melisa1 [442]

Answer:

largest lead = 3 m

Explanation:

Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.

So, first we need to calculate the velocities of both the anchorman  

given data:

Distance = d = 100 m

Time arrival for A = 9.8 s

Time arrival for B = 10.1 s

Velocity of anchorman A = D / Time arrival for A

=100/ 9.8 = 10.2 m/s

Velocity of anchorman B = D / Time arrival for B

=100/10.1 = 9.9 m/s

As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use

d = vt

= 9.9 x 9.8 = 97 m  

So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.

largest lead = 100 - 97 = 3 m

So if his lead no more than 3 m anchorman A win the race.

5 0
3 years ago
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