Question

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Some chipmunks are running back and forth along a tree branch. One chipmunk is at rest at t = 0 s and starts running to the right. It speeds up to a speed of 1.5 m/s in 3 s. How far does it run in this time? Another chipmunk starts from rest and runs to the left. It speeds up to a speed of 2.1 m/s in 3.2 s. It then slows to a stop in 0.5 s and turns around. Where is this squirrel now relative to its starting point? After turning around, the chipmunk in part B runs to the right at a constant speed of 1.8 m/s for 2 s. Draw a velocity vs time and position vs time graph for this chipmunk's motion starting at t = 0

Answer 1

a) u = 0 (starting at rest) ; v = 1.5 m/s ; t = 3 s.

let D be the distance it run.

from equation of motion, v = u + at

a = v/t = 1.5/3 = 0.5 m/s²

Again from third equation, D = ut + 1/2 a t²

D = 0 + 0.5 x 0.5 x 3 x 3 = 2.25 m

Hence, distance covered = D = 2.25 m.

b) u = 0 (this one also starts from rest) ; v1 = 2.1 m/s in t = 3.2 s ; v2 = 0 m/s in t = 0.5 s

We will find the seperate distances for the given motion of the chipmunk and add them to get the total distance.

Using the same way as part (a),

v = u + at

a1 = v/t = 2.1/3.2 = 0.66 m/s²

D1 = 1/2 a t² = 0.5 x 0.66 x 3.2 x 3.2 = 3.38 m (Using D = ut + 1/2 a t²)

a2 = (0 - 2.1) / 0.5 = -4.2 m/s² (Using v = u + at , but u = 2.1 m/s and v = 0 now)

D2 = 2.1 x 0.5 - 0.5 x 4.2 x 0.5 x 0.5 = 0.53 m (Using D = ut + 1/2 a t²)

Total distance from the starting point,

D = D1 + D2 = 3.38 + 0.53 = 3.91 m

Hence, distance covered = D = 3.91 m.

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