A radius vector of point A relative to the origin varies with time t as r → = a t i ^ - b t 2 j ^ , where a and b are constants. Find the equation of the point's trajectory.

The trajectory of a particle is defined as relation between x coordinate and y coordinate of the particle, if it is moving in two-dimensional space. The position vector is a representation of particle's x and y coordinates. Therefore, in our question, we need to eliminate time in order to obtain the relation between coordinates. Given: r → = a t i ^ - b t 2 j ^ Taking the components of this vector into consideration, we get r x = x = a t Finding the value of t from above equation, we get, t = x a and similarly, r y = y = - b t 2 Using the value of t , we get y = - b x a 2 which is the required trajectory.

The speed of a body moving on a straight track varies according to the equation, v = 4 5 t 2 + 3 for 0 < t < 5 s and v = 3 t + 8 for t > 5 s . I f all distances are in metre , find the distance moved by the particle at the end of 10 s .

Velocity can be defined as rate of change of displacement. However, it can be replaced with distance if in the given time interval, velocity does not change its sign. For the interval, 0 < t < 5 , velocity is always positive and for 5 < t < 10 , velocity is again always positive. Therefore, displacement is equal to distance for this case. Given, Velocity is equal to d s d t 0 < t < 5 = 4 5 t 2 + 3 and d s d t t > 5 = 3 t + 8 Then, distance moved in the first 5 s is ⇒ S 1 = ∫ 0 S 1 d s = ∫ 0 5 4 5 t 2 + 3 d t ⇒ S 1 = 4 15 t 3 + 3 t 0 5 = 145 3 m Then, distance moved in the next 5 s is, ⇒ S 2 = ∫ 0 S 2 d s = ∫ 5 10 3 t + 8 d t ⇒ S 2 = 3 t 2 2 + 8 t 5 10 = 305 2 m Total distance moved = S 1 + S 2 = 145 3 + 305 2 m = 1205 6 m