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

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 (t_o, 25)

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! :)

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

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:

120 = \frac12 (t_o)(25)

EDIT: Here's a simple graph:
http://p1cture.me/images/89413308316264556299.png

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."

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

http://p1cture.me/images/00229637359303345669.png

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.