Question

A car starting from rest at A is given an acceleration which increases uniformly

from zero at A to 3.8 m/sec² in 4 seconds. At B, the brakes are applied with constant force and the
car decelerates uniformly by -2.5 m/sec² for 3 seconds until it stops at C. Draw the a-t, v-t and s-t diagram.

Answer 1

To draw the a-t, v-t, and s-t graphs, we calculate the acceleration, velocity, and displacement at different time intervals. The acceleration-time graph will be a straight line starting from (0,0) and ending at (4,3.8). The velocity-time graph will be a straight line starting from (0,0) and ending at (4, 7.6). The displacement-time graph will be a curve with an increasing slope.

Step-by-step explanation:

To draw the graphs, we need to calculate the values of acceleration, velocity, and displacement at different time intervals.

1. Acceleration-time graph (a-t):

The acceleration increases uniformly from zero to 3.8 m/s² in 4 seconds. Therefore, the graph would be a straight line starting from (0,0) and ending at (4,3.8).

2. Velocity-time graph (v-t):

The initial acceleration is 0 m/s² and it increases uniformly to 3.8 m/s² in 4 seconds. The velocity is the area under the acceleration-time graph. It starts from zero at A and increases linearly until it reaches the area of the trapezium under the a-t graph from (0,0) to (4,3.8). The velocity-time graph would be a straight line starting from (0,0) and ending at (4, 7.6).

3. Displacement-time graph (s-t):

The displacement is the area under the velocity-time graph. It starts from zero at A and increases linearly until it reaches the area of the triangle under the v-t graph from (0,0) to (4,7.6).

Therefore, the s-t graph would be a curve with an increasing slope.