Study the given graph and answer the following questions from it. a) Which part of the graph shows accelerated motion? b) Which part shows retardation? Calculate the retardation. c) Calculate the distance travelled by the body in the 1st two hours of journey graphically.

Answers 1

From A to B - it is acceleration, C to D - retardation and distance travelled in first 4s is 8m.

Explanation:

Given graph is a velocity - time graph of an object.

1. In v-t graph from A to B has positive slope, hence this region there is an accelerated motion

acceleration is given by the slope of the curve

a=\frac{\Delta y}{\Delta x}= \frac{4-0}{4-0} =1m/s^{2}a=

Δx

Δy

4−0

=1m/s

Thus, acceleration is 1 m/s^21m/s

2. From C to D, there is a negative slope, which shows the retarded motion.

Slope of the curve is

a=\frac{\Delta y}{\Delta x}= \frac{0-4}{14-12} =-1m/s^{2}a=

Δx

Δy

14−12

0−4

=−1m/s

Thus, retardation is -1 m/s^2

Thus, retardation is -1 m/s^2−1m/s

3. Distance of the body in first 4 seconds is equal to the area under graph from t=0 to t=4s

i.e., area of ∆ABE =\frac{1}{2} b h=\frac{1}{2} \times 4\times 4 = 8m=

bh=

×4×4=8m

Thus, distance travelled by the body in first 4 seconds is 8m