A car has an initial speed of 35 km/h and in 75 seconds it accelerates to 120 km/h. What is the distance travelled during the acceleration period?

A car has an initial speed of 35 km/h and in 75 seconds it accelerates to 120 km/h. What is the distance travelled during the acceleration period?

• A. 2435 m
• B. 1875 m
• C. 1614 m
• D. 1567 m

Answer: (C)

To find the distance traveled during the acceleration period, we first need to calculate the acceleration and then use that to determine the distance. 1. **Calculate acceleration**: The car's initial speed is 35 km/h, which needs to be converted to meters per second for consistency in units. There are 1000 meters in a kilometer and 3600 seconds in an hour, so: \[ \text{Initial speed} = 35 \, \text{km/h} = \frac{35 \times 1000}{3600} \approx 9.72 \, \text{m/s} \] The final speed is 120 km/h, converted similarly: \[ \text{Final speed} = 120 \, \text{km/h} = \frac{120 \times 1000}{3600} \approx 33.33 \, \text{m/s} \] The change in speed (Δv) is: \[ \Delta v = \text{Final speed} - \text{Initial speed} = 33.33 \, \text{m/s} - 9.72 \, \text{

To find the distance traveled during the acceleration period, we first need to calculate the acceleration and then use that to determine the distance.

1. Calculate acceleration: The car's initial speed is 35 km/h, which needs to be converted to meters per second for consistency in units. There are 1000 meters in a kilometer and 3600 seconds in an hour, so:

[

\text{Initial speed} = 35 , \text{km/h} = \frac{35 \times 1000}{3600} \approx 9.72 , \text{m/s}

]

The final speed is 120 km/h, converted similarly:

[

\text{Final speed} = 120 , \text{km/h} = \frac{120 \times 1000}{3600} \approx 33.33 , \text{m/s}

]

The change in speed (Δv) is:

[

\Delta v = \text{Final speed} - \text{Initial speed} = 33.33 , \text{m/s} - 9.72 , \text{