[hicheong]

A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s at a distance of 120 m from its starting point.

1) When the truck has travelled a distance of 60 m from its starting point, its speed is v1 m/s. Which of the following statements concerning v1 is true?

v1 < 12.5 m/s
v1 = 12.5 m/s
v1 > 12.5 m/s

2) When the truck has travelled for a time t2 = t0/2, its distance from its starting point is s2. Which of the following statements concerning s2 is true?

s2 < 60 m
s2 = 60 m
s2 > 60 m

[Unknown008]

1. It would really help you if you could draw the graph of the motion, more specifically, the velocity-time graph.

In this graph, the gradient gives the acceleration.
You therefore have a straight line passing through zero and going up with constant gradient.

On the time axis, put t0 somewhere. This time corresponds to a speed of 25 m/s, that is, you have the coordinates

The area under the graph gives the distance travelled.

So, the area of the triangle formed is 120.

If you observe well, for the car to travel 60 m (that is half the area), it must have been going for more than (t0)/2 seconds. But at (t0)/2 seconds, the speed is 25/2 = 12.5 m/s. Since the car travelled for more time than this, the speed reached then should be?

2. You apply the same logic as previously.

I will repeat again just in case. You should draw a velocity-time graph. This will help you a lot.

Post what you get! :)

[hicheong]

A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s at a distance of 120 m from its starting point.

How long does it take for the pickup to reach its speed of 25 m/s?

t0 = 3.1 s
t0 = 4.8 s
t0 = 6.8 s
t0 = 9.6 s
t0 = 13.4 s

[Unknown008]

As I told you before, you should draw a velocity time graph. You'll see that the area (which is also the distance) of the triangle formed will given by:



EDIT: Here's a simple graph:

[hicheong]

But how do you come up with the statement that "If you observe well, for the car to travel 60 m (that is half the area), it must have been going for more than (t0)/2 seconds."

[Unknown008]

Hm... you see the graph I posted, I'll add some more colours.



You you see that the reddish area is smaller than the blueish area?

Red + blue = 120 m

For the smaller triangle (red) to have equal area to the right trapezium (blue), the base of the triangle should be larger, and the width of the trapezium smaller.