Q 3.27  The speed-time graph of a particle moving along a fixed direction is shown in Figure. Obtain the distance traversed by the particle between

(b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)?

Answers (1)
S Sayak

As the speed is increasing in the time interval t = 0 s to t = 5 s the acceleration is positive and can be given by

\\a_{1}=\frac{12}{5}\\ a_{1}=2.4ms^{-2}

Speed at t = 2 s is

\\u_{1}=0+2.4\times 2\\ u_{1}=4.8ms^{-1}

Speed at t = 5 s is v1 = 15 m s-1 

t= 5 - 2 = 3 s 

Distance travelled in interval t = 2 s to t = 5 s is s1

\\s_{1}=u_{1}t_{1}+\frac{1}{2}a_{1}t_{1}^{2}\\ s_{1}=4.8\times 3+1.2\times 3^{2}\\ s_{1}=25.2m

Acceleration is negative after t = 5 s but has the same magnitude

a= -2.4 m s-2

Speed at t =  5 s is u2 = 12 m s-1

t= 6 - 5 = 1 s

Distance travelled in this interval can be calculated as follows

\\s_{2}=u_{2}t_{2}+\frac{1}{2}a_{2}t_{2}^{2}\\ s_{2}=12\times 1-1.2\times 1^{2}\\ s_{2}=10.8m

Total distance travelled from t = 2 s to t = 6 s is s = s1 + s2

s=25.2+10.8

s=36 m

The average speed over this interval is 

\\v_{avg}=\frac{s}{t_{1}+t_{2}}\\ v_{avg}=\frac{36}{3+1}\\ v_{avg}=9\ ms^{-1}

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