David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0m/s2at the instant when David passes. How far does Tina drive before passing David?

Short Answer

Expert verified

The kinematic equations of motion at constant acceleration is given by:

vfs=vis+asΔtsf=si+vsΔt+12asΔt2vfs2=vis2+2asΔs

Tina drives 900 m before passing David.

Step by step solution

01

Draw the pictorial diagram and the motion diagram:

The following steps have to be taken for the pictorial presentation:

Draw a motion diagram.

Establish coordinates.

Sketch the situation.

Define symbols.

List knowns.

Identify desired unknown.

02

Calculating the distance Tina drove before passing David:

The velocity at which the David is travelling is given by v1=30 m/s.

The distance travelled is given by s=30t where t is the time taken.

The time taken for David to travel when he passes Tina is given by the kinematic equation of motion:

s=ut+12at230t=0+122t2t=0sor30s

The velocity is defined as the rate of change of displacement with respect to time.

The distance traveled by Tina is given by :

s=0+12×2m/s2×302=900m

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