Question

The speed of a body moving on a straight track varies according to v = 4t +1 for t> 7s. The distances
are measured in metre. The distance measured in metres moved by the particle at the end of 10 second is
1) 127
2) 247
3) 186
4) 313
​

Answer 1

Answer:

Total distance moved by the body is 247 m

Explanation:

As we know that the velocity of the particle at given interval of time given in this question

So for the first interval of time we have

$v = 2t + 13$ from t = 0 to t = 5 s

so we have

$s_1 = \int (2t + 13) dt$

$s_1 = (t^2 + 13 t)$

time limits are from 0 to 5 s so we have

$s_1 = 25 + 65 = 90 m$

now for next interval of time the speed is given as

$v = 3t + 8$ from t = 5 to t = 7

so we have

$s_2 = \int (3t + 8) dt$

$s_2 = 1.5 t^2 + 8 t$

now limits are t = 5 to t = 7 s

$s_2 = 1.5 (49 - 25) + 8 (7 - 5)$

$s_2 = 52 m$

now for last interval of time speed is given as

$v = 4t + 1$

so the distance moved is given as

$s_3 = \int (4t + 1) dt$

$s_3 = 2t^2 + t$

limits are t = 7 to t = 10 s

$s_3 = 2(100 - 49) + (10 - 7)$

$s_3 = 105 m$

So total distance moved by it is given as

$d = 90 + 52 + 105$

$d = 247 m$

#Learn

Topic : Kinematics

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