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An antelope moving with constant acceleration covers the distance between two points 70.0 \(\mathrm{m}\) apart in 7.00 \(\mathrm{s} .\) Its speed as it passes the second point is 15.0 \(\mathrm{m} / \mathrm{s}\) . (a) What is its speed at the first point? (b) What is its acceleration?

Short Answer

Expert verified
The initial speed is 5.0 m/s and the acceleration is approximately 1.43 m/s².

Step by step solution

01

Define known values

The distance between two points is given as \( s = 70.0 \, \text{m} \), the time taken to cover this distance is \( t = 7.00 \, \text{s} \), and the final speed at the second point is \( v = 15.0 \, \text{m/s} \).
02

Use the kinematic equation for distance

The kinematic equation \( s = ut + \frac{1}{2} a t^2 \) relates the initial speed \( u \), acceleration \( a \), time \( t \), and distance \( s \). We will use this equation along with the given values, acknowledging that we have two unknowns: \( u \) and \( a \).
03

Use another kinematic equation for speed

Use the equation \( v = u + at \) where \( v = 15.0 \, \text{m/s} \) is the final speed. We will use this to express \( a \) in terms of \( u \): \( a = \frac{v-u}{t} \).
04

Substitute for acceleration

Substitute \( a = \frac{v-u}{t} \) from Step 3 into the equation from Step 2: \[ 70.0 = u \times 7.00 + \frac{1}{2} \left( \frac{15.0 - u}{7.00} \right) \times 7.00^2 \].
05

Solve for initial speed (u)

After simplifying the equation from Step 4, solve for \( u \): \[ 70.0 = 7u + \frac{1}{2} ((15 - u) \times 7) \]. Simplifying further, \[ 70.0 = 7u + 52.5 - 3.5u \]. Therefore, \[ 70.0 = 3.5u + 52.5 \]. Solving this gives \( 3.5u = 17.5 \) implying \( u = 5.0 \, \text{m/s} \).
06

Solve for acceleration (a)

Using the value of \( u = 5.0 \, \text{m/s} \) from Step 5 and the equation \( a = \frac{v-u}{t} \): \[ a = \frac{15.0 - 5.0}{7.0} = \frac{10.0}{7.0} \]. Therefore, \( a \approx 1.43 \, \text{m/s}^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform acceleration
Uniform acceleration refers to a constant rate of change in velocity over time. In simpler terms, if an object is accelerating uniformly, its speed is increasing by the same amount every second. This makes calculations straightforward because once you know the initial speed and the acceleration, you can predict the object's future speed and position.

In the antelope problem, the animal is moving under uniform acceleration between two points. This means that its acceleration does not change as it travels the 70 meters. Knowing this helps us use kinematic equations to calculate other variables such as its initial speed. Uniform acceleration is a common assumption in physics problems that allows us to use some handy equations for further analysis.
Kinematic equations
Kinematic equations are tools physicists and engineers use to analyze the motion of objects under uniform acceleration. These equations help find unknown variables when others are known, such as distance, time, initial speed, final speed, and acceleration. For the given problem, we use two primary kinematic equations.

  • The first equation, \( s = ut + \frac{1}{2} a t^2 \), is used to relate initial velocity \( u \), time \( t \), distance \( s \), and acceleration \( a \). This equation helps us determine how far an object travels under constant acceleration.
  • The second key equation, \( v = u + at \), connects final speed \( v \), initial speed \( u \), acceleration \( a \), and time \( t \). It tells us how the velocity changes over time.
These equations are crucial in connecting different aspects of the motion in the problem. By substituting known values into these equations, we can methodically solve for the unknowns, as demonstrated in the solution steps.
Initial velocity
Initial velocity is the speed of an object before it begins to accelerate. It's an essential starting point in many physics problems, such as the antelope's motion. When acceleration is uniform, we can calculate or deduce the initial velocity using kinematic equations.

In the antelope's case, we solved for its initial velocity using the data provided. We knew the final speed at the second point was 15 m/s, the time taken was 7 seconds, and the distance covered was 70 meters. With these details, we used the kinematic equation formulas to rearrange and simplify, eventually calculating that the initial velocity was 5 m/s.

Understanding initial velocity is important in predicting how an object will move, providing a foundation for further analysis involving speeds and distances in kinematic studies.

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Most popular questions from this chapter

A 15-kg rock is dropped from rest on the earth and reaches the ground in 1.75 \(\mathrm{s}.\) When it is dropped from the same height on Saturn's satellite Enceladus, it reaches the ground in 18.6 \(\mathrm{s}.\) What is the acceleration due to gravity on Enceladus?

Large cockroaches can run as fast as 1.50 \(\mathrm{m} / \mathrm{s}\) in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant 1.50 \(\mathrm{m} / \mathrm{s} .\) If you start 0.90 m behind the cockroach with an initial speed of 0.80 \(\mathrm{m} / \mathrm{s}\) toward it, what minimum constant acceleration would you need to catch up with it when it has traveled \(1.20 \mathrm{m},\) just short of safety under a counter?

Passing. The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 \(\mathrm{m} / \mathrm{s}\) (about 45 \(\mathrm{mi} / \mathrm{h} )\) . Initially, the car is also traveling at 20.0 \(\mathrm{m} / \mathrm{s}\) and its front bumper is 24.0 \(\mathrm{m}\) behind the truck's rear bumper. The car accelerates at a constant 0.600 \(\mathrm{m} / \mathrm{s}^{2}\) , then pulls back into the truck's lane when the rear of the car is 26.0 \(\mathrm{m}\) ahead of the front of the truck. The car is 4.5 \(\mathrm{m}\) long and the truck is 21.0 \(\mathrm{m}\) long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?

A Honda Civic travels in a straight line along a road. Its distance \(x\) from a stop sign is given as a function of time \(t\) by the equation \(x(t)=\alpha t^{2}-\beta t^{3},\) where \(\alpha=1.50 \mathrm{m} / \mathrm{s}^{2}\) and \(\beta=\) 0.0500 \(\mathrm{m} / \mathrm{s}^{3} .\) Calculate the average velocity of the car for each time interval: \((\mathrm{a}) t=0\) to \(t=2.00 \mathrm{s} ;\) (b) \(t=0\) to \(t=4.00 \mathrm{s}\) ; (c) \(t=2.00\) s to \(t=4.00 \mathrm{s}.\)

A jet fighter pilot wishes to accelerate from rest at a constant acceleration of 5\(g\) to reach Mach 3 (three times the speed of sound) as quickly as possible. Experimental tests reveal that he will black out if this acceleration lasts for more than 5.0 s. Use 331 \(\mathrm{m} / \mathrm{s}\) for the speed of sound. (a) Will the period of acceleration last long enough to cause him to black out? (b) What is the greatest speed he can reach with an acceleration of 5\(g\) before blacking out?

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