Question
a car is travelling at 20m/s along a straight line. the brakes are applied and the car retardation is 3m/s2 for 5s. find the cars final velocity
Answers
GPT-4o
To find the car's final velocity after the brakes are applied, we can use one of the basic kinematic equations for uniformly accelerated (or decelerated) motion. The equation we'll use is:
\[ v_f = v_i + a t \]
where:
- \( v_f \) is the final velocity,
- \( v_i \) is the initial velocity,
- \( a \) is the acceleration (or in this case, deceleration since it is negative),
- \( t \) is the time.
Given:
- The initial velocity (\( v_i \)) is 20 m/s,
- The retardation (deceleration, \( a \)) is \(-3 \) m/s² (it's negative because it's a deceleration),
- The time (\( t \)) is 5 seconds.
Plugging in these values:
\[ v_f = 20 \text{ m/s} + (-3 \text{ m/s}^2 \times 5 \text{ s}) \]
\[ v_f = 20 \text{ m/s} - 15 \text{ m/s} \]
\[ v_f = 5 \text{ m/s} \]
So the car's final velocity after applying the brakes for 5 seconds is 5 m/s.
\[ v_f = v_i + a t \]
where:
- \( v_f \) is the final velocity,
- \( v_i \) is the initial velocity,
- \( a \) is the acceleration (or in this case, deceleration since it is negative),
- \( t \) is the time.
Given:
- The initial velocity (\( v_i \)) is 20 m/s,
- The retardation (deceleration, \( a \)) is \(-3 \) m/s² (it's negative because it's a deceleration),
- The time (\( t \)) is 5 seconds.
Plugging in these values:
\[ v_f = 20 \text{ m/s} + (-3 \text{ m/s}^2 \times 5 \text{ s}) \]
\[ v_f = 20 \text{ m/s} - 15 \text{ m/s} \]
\[ v_f = 5 \text{ m/s} \]
So the car's final velocity after applying the brakes for 5 seconds is 5 m/s.