Question

a car with its brakes applied has an acceleration of -1.2m/s squared. if its initial velocity is 10m/s, the distance the car covers in the first 5 seconds after the acceleration begins is

Answer 1

To calculate the distance the car covers in the first 5 seconds after the acceleration begins, we can use the following equation of motion:

\[
d = v_0 t + \frac{1}{2} a t^2
\]

where:
- \( d \) is the distance covered,
- \( v_0 \) is the initial velocity,
- \( t \) is the time,
- \( a \) is the acceleration.

In this case:
- \( v_0 = 10 \, \text{m/s} \) (initial velocity),
- \( a = -1.2 \, \text{m/s}^2 \) (acceleration),
- \( t = 5 \, \text{s} \) (time).

Now, substitute the values into the equation:

\[
d = (10 \, \text{m/s}) \cdot (5 \, \text{s}) + \frac{1}{2} \cdot (-1.2 \, \text{m/s}^2) \cdot (5 \, \text{s})^2
\]

Calculating each part:

1. The first term:
\[
(10 \, \text{m/s}) \cdot (5 \, \text{s}) = 50 \, \text{m}
\]

2. The second term:
\[
\frac{1}{2} \cdot (-1.2 \, \text{m/s}^2) \cdot 25 \, \text{s}^2 = \frac{1}{2} \cdot (-1.2) \cdot 25 = -15 \, \text{m}
\]

Now, combine these two results:

\[
d = 50 \, \text{m} - 15 \, \text{m} = 35 \, \text{m}
\]

So, the distance the car covers in the first 5 seconds after the acceleration begins is **35 meters**.