Question

A body start from rest and moves in a straight path with uniform acceleration of 8m/s² for 5 secs. It then decelerates uniformly to rest in the next 10 secs. calculate the magnitude of its deceleration

Answer 1

To solve this problem, we first need to find out how fast the body is moving at the end of the first segment of motion (the acceleration phase) and then determine the deceleration during the second segment (the deceleration phase).

1. **First Segment (Uniform Acceleration):**
- Initial velocity (\(u\)) = 0 m/s (starts from rest)
- Acceleration (\(a\)) = 8 m/s²
- Time (\(t\)) = 5 seconds

Using the formula for final velocity:
\[
v = u + at
\]
we can calculate the final velocity (\(v\)) at the end of the first 5 seconds:
\[
v = 0 + (8 \, \text{m/s}^2) \times (5 \, \text{s}) = 40 \, \text{m/s}
\]

2. **Second Segment (Uniform Deceleration):**
- Initial velocity for this phase (\(v_i\)) = 40 m/s (the final velocity from the first segment)
- Final velocity (\(v_f\)) = 0 m/s (comes to rest)
- Time (\(t\)) = 10 seconds

Using the formula for uniform acceleration (which can be used for deceleration as well):
\[
v_f = v_i + a_d t
\]
where \(a_d\) is the deceleration (which will be negative).

Plugging in the values:
\[
0 = 40 + a_d \times 10
\]

Solving for \(a_d\):
\[
a_d \times 10 = -40
\]
\[
a_d = -4 \, \text{m/s}^2
\]

The magnitude of the deceleration is:
\[
|a_d| = 4 \, \text{m/s}^2
\]

Thus, the magnitude of the deceleration is **4 m/s²**.