In testing an 800-kg sports car it is found that the engine rotates the tires, indirectly causing a forward pushing force, F on the tires, during the first 10 seconds as shown in the figure. 5000 0 F (N) 10 t(s) Ignore friction and air drag forces and assume that this force F is the only force acting on the car in the direction of motion. If the car starts from rest, what is the its speed after the 10 seconds time interval measured in m/s?
In testing an 800-kg sports car it is found that the engine rotates the tires, indirectly causing a forward pushing force, F on the tires, during the first 10 seconds as shown in the figure. 5000 0 F (N) 10 t(s) Ignore friction and air drag forces and assume that this force F is the only force acting on the car in the direction of motion. If the car starts from rest, what is the its speed after the 10 seconds time interval measured in m/s?
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[Description of the Graph]
- The graph showcases a Force (\( F \)) versus Time (\( t \)) relationship.
- The x-axis represents time, \( t \), in seconds (s), ranging from 0 to 10 seconds.
- The y-axis represents force, \( F \), in Newtons (N), ranging from 0 to 5000 N.
- The plot forms a triangle, starting at the origin (0, 0), increasing linearly to the peak at \( F = 5000 \) N and \( t = 5 \) s, and then decreasing linearly back to 0 N at \( t = 10 \) s.
### Calculation
Ignore friction and air drag forces and assume that this force \( F \) is the only force acting on the car in the direction of motion. If the car starts from rest, what is its speed after the 10 seconds time interval measured in m/s?
### Solution:
First, we need to determine the acceleration since the force \( F \) changes over time. According to Newton's second law:
\[ F = ma \]
where \( m \) is the mass of the car and \( a \) is the acceleration. Here, we need to find the net impulse delivered to the car during the 10 seconds, which is the area under the force-time graph.
1. **Calculate the area under the graph**:
Since the force-time graph forms a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
\[
\text{Base} = 10 \, \text{s}
\]
\[
\text{Height} = 5000 \, \text{N}
\]
\[
\text{Area} = \frac{1}{2} \times 10 \times 5000 = 25000 \, \text{Ns}
\]
The area under the force-time graph gives the impulse,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcde4937d-e883-446c-8941-8d2ade4d99b0%2F9db086a8-2bd1-4c3a-9186-98feb14108b9%2Fmluaftr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
In testing an 800-kg sports car, it is found that the engine rotates the tires, indirectly causing a forward pushing force, \( F \) on the tires, during the first 10 seconds as shown in the figure below.
[Description of the Graph]
- The graph showcases a Force (\( F \)) versus Time (\( t \)) relationship.
- The x-axis represents time, \( t \), in seconds (s), ranging from 0 to 10 seconds.
- The y-axis represents force, \( F \), in Newtons (N), ranging from 0 to 5000 N.
- The plot forms a triangle, starting at the origin (0, 0), increasing linearly to the peak at \( F = 5000 \) N and \( t = 5 \) s, and then decreasing linearly back to 0 N at \( t = 10 \) s.
### Calculation
Ignore friction and air drag forces and assume that this force \( F \) is the only force acting on the car in the direction of motion. If the car starts from rest, what is its speed after the 10 seconds time interval measured in m/s?
### Solution:
First, we need to determine the acceleration since the force \( F \) changes over time. According to Newton's second law:
\[ F = ma \]
where \( m \) is the mass of the car and \( a \) is the acceleration. Here, we need to find the net impulse delivered to the car during the 10 seconds, which is the area under the force-time graph.
1. **Calculate the area under the graph**:
Since the force-time graph forms a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
\[
\text{Base} = 10 \, \text{s}
\]
\[
\text{Height} = 5000 \, \text{N}
\]
\[
\text{Area} = \frac{1}{2} \times 10 \times 5000 = 25000 \, \text{Ns}
\]
The area under the force-time graph gives the impulse,
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