# A car speeding up from rest to constant accel. travels 130m in 6s. How far in

• Butterfly30
In summary, a car accelerating from rest to a constant speed of 16.5m/s travels 100 meters in 6.04 seconds. In order to find the distance travelled in the first 2.37 seconds, the formula d=vt is used, resulting in a distance of 37.03 meters. However, this is incorrect and it is suggested to instead draw a graph of velocity vs. time to find the slope and y-intercept. This will allow for more accurate calculations as the numbers involved are high and have decimals.

#### Butterfly30

A car speeding up from rest to constant accel. travels 100 meters in 6.04s. How far did it travel in the first 2.37 seconds?

So I figure if I divide distance by time 100/6.04 it will give me the velocity 16.5m/s. And since distance = vt I would do d=15.6*2.37 which equals 37.03m but this is wrong. I was also told to try drawing a graph of velocity vs time...but how can I graph numbers such as these (high and low #'s with decimals) If I can't graph it I can't find the slope or y intercept...so I'm lost

Last edited:
Draw a graph of velocity vs. time.
Distance travelled=velocity x time

## 1. How do you determine the constant acceleration of the car?

The constant acceleration of the car can be determined using the equation a = (v - u)/t, where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time taken. In this case, the initial velocity is 0 m/s, the final velocity is unknown, and the time taken is 6 seconds. Therefore, the equation becomes a = v/6. To find the value of acceleration, we need to know the final velocity of the car.

## 2. How can we calculate the final velocity of the car?

The final velocity of the car can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. In this case, the initial velocity is 0 m/s, the acceleration is calculated in the previous step, and the time taken is 6 seconds. Therefore, the equation becomes v = at. Once we know the value of the final velocity, we can calculate the distance traveled by the car.

## 3. What is the distance traveled by the car?

The distance traveled by the car can be calculated using the equation s = ut + 1/2 at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken. In this case, the initial velocity is 0 m/s, the acceleration is calculated in the first step, and the time taken is 6 seconds. Therefore, the equation becomes s = 1/2 at^2. By substituting the values, we get s = 1/2 (a)(6)^2 = 3a = 3(v/6) = 0.5v. Therefore, the distance traveled by the car is 0.5 times the final velocity.

## 4. Can we calculate the final velocity if the time taken is not given?

Yes, we can calculate the final velocity using the equation v = √(u^2 + 2as), where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled. In this case, the initial velocity is 0 m/s, the acceleration is calculated in the first step, and the distance traveled is 130m. Therefore, the equation becomes v = √(0 + 2a(130)) = √(260a). Since we know the value of acceleration, we can calculate the final velocity.

## 5. Is the acceleration of the car constant throughout the motion?

Yes, the acceleration of the car is constant throughout the motion as per the given information. This is because the car starts from rest and reaches a constant speed in 6 seconds, which means that the acceleration remains the same throughout the motion.