Question

a 6 kg cart initiakky traveling at 4ms east accelerates uniformly at 0.50 Ms quared east for 3 seconds what's the speed of the cart at the end of the 3 seconds interval

Answer 1

To calculate the final speed of the cart after 3 seconds of uniform acceleration, you can use the formula for final velocity:

\[ v_f = v_i + a \cdot t \]

Where:
- \( v_f \) = final velocity
- \( v_i \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

Given:
- \( v_i = 4 \, \text{m/s} \) (east)
- \( a = 0.50 \, \text{m/s}^2 \) (east)
- \( t = 3 \, \text{s} \)

Now, plug in the values:

\[
v_f = 4 \, \text{m/s} + (0.50 \, \text{m/s}^2 \cdot 3 \, \text{s})
\]

Calculating the acceleration term:

\[
0.50 \, \text{m/s}^2 \cdot 3 \, \text{s} = 1.5 \, \text{m/s}
\]

Now substitute this back into the equation for \( v_f \):

\[
v_f = 4 \, \text{m/s} + 1.5 \, \text{m/s} = 5.5 \, \text{m/s}
\]

Thus, the speed of the cart at the end of the 3 seconds interval is \( \mathbf{5.5 \, m/s} \) east.